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• Number guessing game (3+- guessed result)

  - by Nick Waring

  I've been assigned a task to create a game that generates 4 digits and the user has to guess the digits one at a time to get the correct result. If the number is correct a Y is displayed and if not, a N. This was easy, now the next step was to implement another two responses. If the answer is too high, a H is displayed and too low, an N. Again, was easy - now the third is to use the same design as game 2 but if the number is 3 higher than a H is displayed and same if it's 3 lower than a L is displayed - otherwise an X is displayed. I can't figure out how to do this. Here's my test code for game 2 for just one of the digits - any help is appreciated. (5 was used just for a test.) def guess(): x = 5 g= int(input("Guess the number: ")) if g == x: print("Y") elif g < x: print("L") else: print("H")

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• How to set up a specific registration on Database

  - by ymorenz

  I'm creating a database so I can access names of people from a university. But I would like to create a registration ID for each person like 1012607, of which the first 2 numbers would be the year (11 for 2011) and the third digit would be the semester they registered (1 in the beginning of the year and 2 in the end of the year). The other 4 digits would be incremental. I don't know how to code a prefixed number in MySQL that can change only the last 4 digits and after a year change the first 2 digits every time you have a new registration. Can someone please help me?

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• How to optimize dynamic programming?

  - by Chan

  Problem A number is called lucky if the sum of its digits, as well as the sum of the squares of its digits is a prime number. How many numbers between A and B are lucky? Input: The first line contains the number of test cases T. Each of the next T lines contains two integers, A and B. Output: Output T lines, one for each case containing the required answer for the corresponding case. Constraints: 1 <= T <= 10000 1 <= A <= B <= 10^18 Sample Input: 2 1 20 120 130 Sample Output: 4 1 Explanation: For the first case, the lucky numbers are 11, 12, 14, 16. For the second case, the only lucky number is 120. The problem is quite simple if we use brute force, however the running time is so critical that my program failed most test cases. My current idea is to use dynamic programming by storing the previous sum in a temporary array, so for example: sum_digits(10) = 1 -> sum_digits(11) = sum_digits(10) + 1 The same idea is applied for sum square but with counter equals to odd numbers. Unfortunately, it still failed 9 of 10 test cases which makes me think there must be a better way to solve it. Any idea would be greatly appreciated. #include <iostream> #include <vector> #include <string> #include <algorithm> #include <unordered_map> #include <unordered_set> #include <cmath> #include <cassert> #include <bitset> using namespace std; bool prime_table[1540] = { 0, 0, 1, 1, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0 }; unsigned num_digits(long long i) { return i > 0 ? (long) log10 ((double) i) + 1 : 1; } void get_sum_and_sum_square_digits(long long n, int& sum, int& sum_square) { sum = 0; sum_square = 0; int digit; while (n) { digit = n % 10; sum += digit; sum_square += digit * digit; n /= 10; } } void init_digits(long long n, long long previous_sum[], const int size = 18) { int current_no_digits = num_digits(n); int digit; for (int i = 0; i < current_no_digits; ++i) { digit = n % 10; previous_sum[i] = digit; n /= 10; } for (int i = current_no_digits; i <= size; ++i) { previous_sum[i] = 0; } } void display_previous(long long previous[]) { for (int i = 0; i < 18; ++i) { cout << previous[i] << ","; } } int count_lucky_number(long long A, long long B) { long long n = A; long long end = B; int sum = 0; int sum_square = 0; int lucky_counter = 0; get_sum_and_sum_square_digits(n, sum, sum_square); long long sum_counter = sum; long long sum_square_counter = sum_square; if (prime_table[sum_counter] && prime_table[sum_square_counter]) { lucky_counter++; } long long previous_sum[19] = {1}; init_digits(n, previous_sum); while (n < end) { n++; if (n % 100000000000000000 == 0) { previous_sum[17]++; sum_counter = previous_sum[17] + previous_sum[18]; sum_square_counter = previous_sum[17] * previous_sum[17] + previous_sum[18] * previous_sum[18]; previous_sum[16] = 0; previous_sum[15] = 0; previous_sum[14] = 0; previous_sum[13] = 0; previous_sum[12] = 0; previous_sum[11] = 0; previous_sum[10] = 0; previous_sum[9] = 0; previous_sum[8] = 0; previous_sum[7] = 0; previous_sum[6] = 0; previous_sum[5] = 0; previous_sum[4] = 0; previous_sum[3] = 0; previous_sum[2] = 0; previous_sum[1] = 0; previous_sum[0] = 0; } else if (n % 10000000000000000 == 0) { previous_sum[16]++; sum_counter = previous_sum[16] + previous_sum[17] + previous_sum[18]; sum_square_counter = previous_sum[16] * previous_sum[16] + previous_sum[17] * previous_sum[17] + previous_sum[18] * previous_sum[18]; previous_sum[15] = 0; previous_sum[14] = 0; previous_sum[13] = 0; previous_sum[12] = 0; previous_sum[11] = 0; previous_sum[10] = 0; previous_sum[9] = 0; previous_sum[8] = 0; previous_sum[7] = 0; previous_sum[6] = 0; previous_sum[5] = 0; previous_sum[4] = 0; previous_sum[3] = 0; previous_sum[2] = 0; previous_sum[1] = 0; previous_sum[0] = 0; } else if (n % 1000000000000000 == 0) { previous_sum[15]++; sum_counter = previous_sum[15] + previous_sum[16] + previous_sum[17] + previous_sum[18]; sum_square_counter = previous_sum[15] * previous_sum[15] + previous_sum[16] * previous_sum[16] + previous_sum[17] * previous_sum[17] + previous_sum[18] * previous_sum[18]; previous_sum[14] = 0; previous_sum[13] = 0; previous_sum[12] = 0; previous_sum[11] = 0; previous_sum[10] = 0; previous_sum[9] = 0; previous_sum[8] = 0; previous_sum[7] = 0; previous_sum[6] = 0; previous_sum[5] = 0; previous_sum[4] = 0; previous_sum[3] = 0; previous_sum[2] = 0; previous_sum[1] = 0; previous_sum[0] = 0; } else if (n % 100000000000000 == 0) { previous_sum[14]++; sum_counter = previous_sum[14] + previous_sum[15] + previous_sum[16] + previous_sum[17] + previous_sum[18]; sum_square_counter = previous_sum[14] * previous_sum[14] + previous_sum[15] * previous_sum[15] + previous_sum[16] * previous_sum[16] + previous_sum[17] * previous_sum[17] + previous_sum[18] * previous_sum[18]; previous_sum[13] = 0; previous_sum[12] = 0; previous_sum[11] = 0; previous_sum[10] = 0; previous_sum[9] = 0; previous_sum[8] = 0; previous_sum[7] = 0; previous_sum[6] = 0; previous_sum[5] = 0; previous_sum[4] = 0; previous_sum[3] = 0; previous_sum[2] = 0; previous_sum[1] = 0; previous_sum[0] = 0; } else if (n % 10000000000000 == 0) { previous_sum[13]++; sum_counter = previous_sum[13] + previous_sum[14] + previous_sum[15] + previous_sum[16] + previous_sum[17] + previous_sum[18]; sum_square_counter = previous_sum[13] * previous_sum[13] + previous_sum[14] * previous_sum[14] + previous_sum[15] * previous_sum[15] + previous_sum[16] * previous_sum[16] + previous_sum[17] * previous_sum[17] + previous_sum[18] * previous_sum[18]; previous_sum[12] = 0; previous_sum[11] = 0; previous_sum[10] = 0; previous_sum[9] = 0; previous_sum[8] = 0; previous_sum[7] = 0; previous_sum[6] = 0; previous_sum[5] = 0; previous_sum[4] = 0; previous_sum[3] = 0; previous_sum[2] = 0; previous_sum[1] = 0; previous_sum[0] = 0; } else if (n % 1000000000000 == 0) { previous_sum[12]++; sum_counter = previous_sum[12] + previous_sum[13] + previous_sum[14] + previous_sum[15] + previous_sum[16] + previous_sum[17] + previous_sum[18]; sum_square_counter = previous_sum[12] * previous_sum[12] + previous_sum[13] * previous_sum[13] + previous_sum[14] * previous_sum[14] + previous_sum[15] * previous_sum[15] + previous_sum[16] * previous_sum[16] + previous_sum[17] * previous_sum[17] + previous_sum[18] * previous_sum[18]; previous_sum[11] = 0; previous_sum[10] = 0; previous_sum[9] = 0; previous_sum[8] = 0; previous_sum[7] = 0; previous_sum[6] = 0; previous_sum[5] = 0; previous_sum[4] = 0; previous_sum[3] = 0; previous_sum[2] = 0; previous_sum[1] = 0; previous_sum[0] = 0; } else if (n % 100000000000 == 0) { previous_sum[11]++; sum_counter = previous_sum[11] + previous_sum[12] + previous_sum[13] + previous_sum[14] + previous_sum[15] + previous_sum[16] + previous_sum[17] + previous_sum[18]; sum_square_counter = previous_sum[11] * previous_sum[11] + previous_sum[12] * previous_sum[12] + previous_sum[13] * previous_sum[13] + previous_sum[14] * previous_sum[14] + previous_sum[15] * previous_sum[15] + previous_sum[16] * previous_sum[16] + previous_sum[17] * previous_sum[17] + previous_sum[18] * previous_sum[18]; previous_sum[10] = 0; previous_sum[9] = 0; previous_sum[8] = 0; previous_sum[7] = 0; previous_sum[6] = 0; previous_sum[5] = 0; previous_sum[4] = 0; previous_sum[3] = 0; previous_sum[2] = 0; previous_sum[1] = 0; previous_sum[0] = 0; } else if (n % 10000000000 == 0) { previous_sum[10]++; sum_counter = previous_sum[10] + previous_sum[11] + previous_sum[12] + previous_sum[13] + previous_sum[14] + previous_sum[15] + previous_sum[16] + previous_sum[17] + previous_sum[18]; sum_square_counter = previous_sum[10] * previous_sum[10] + previous_sum[11] * previous_sum[11] + previous_sum[12] * previous_sum[12] + previous_sum[13] * previous_sum[13] + previous_sum[14] * previous_sum[14] + previous_sum[15] * previous_sum[15] + previous_sum[16] * previous_sum[16] + previous_sum[17] * previous_sum[17] + previous_sum[18] * previous_sum[18]; previous_sum[9] = 0; previous_sum[8] = 0; previous_sum[7] = 0; previous_sum[6] = 0; previous_sum[5] = 0; previous_sum[4] = 0; previous_sum[3] = 0; previous_sum[2] = 0; previous_sum[1] = 0; previous_sum[0] = 0; } else if (n % 1000000000 == 0) { previous_sum[9]++; sum_counter = previous_sum[9] + previous_sum[10] + previous_sum[11] + previous_sum[12] + previous_sum[13] + previous_sum[14] + previous_sum[15] + previous_sum[16] + previous_sum[17] + previous_sum[18]; sum_square_counter = previous_sum[9] * previous_sum[9] + previous_sum[10] * previous_sum[10] + previous_sum[11] * previous_sum[11] + previous_sum[12] * previous_sum[12] + previous_sum[13] * previous_sum[13] + previous_sum[14] * previous_sum[14] + previous_sum[15] * previous_sum[15] + previous_sum[16] * previous_sum[16] + previous_sum[17] * previous_sum[17] + previous_sum[18] * previous_sum[18]; previous_sum[8] = 0; previous_sum[7] = 0; previous_sum[6] = 0; previous_sum[5] = 0; previous_sum[4] = 0; previous_sum[3] = 0; previous_sum[2] = 0; previous_sum[1] = 0; previous_sum[0] = 0; } else if (n % 100000000 == 0) { previous_sum[8]++; sum_counter = previous_sum[8] + previous_sum[9] + previous_sum[10] + previous_sum[11] + previous_sum[12] + previous_sum[13] + previous_sum[14] + previous_sum[15] + previous_sum[16] + previous_sum[17] + previous_sum[18]; sum_square_counter = previous_sum[8] * previous_sum[8] + previous_sum[9] * previous_sum[9] + previous_sum[10] * previous_sum[10] + previous_sum[11] * previous_sum[11] + previous_sum[12] * previous_sum[12] + previous_sum[13] * previous_sum[13] + previous_sum[14] * previous_sum[14] + previous_sum[15] * previous_sum[15] + previous_sum[16] * previous_sum[16] + previous_sum[17] * previous_sum[17] + previous_sum[18] * previous_sum[18]; previous_sum[7] = 0; previous_sum[6] = 0; previous_sum[5] = 0; previous_sum[4] = 0; previous_sum[3] = 0; previous_sum[2] = 0; previous_sum[1] = 0; previous_sum[0] = 0; } else if (n % 10000000 == 0) { previous_sum[7]++; sum_counter = previous_sum[7] + previous_sum[8] + previous_sum[9] + previous_sum[10] + previous_sum[11] + previous_sum[12] + previous_sum[13] + previous_sum[14] + previous_sum[15] + previous_sum[16] + previous_sum[17] + previous_sum[18]; sum_square_counter = previous_sum[7] * previous_sum[7] + previous_sum[8] * previous_sum[8] + previous_sum[9] * previous_sum[9] + previous_sum[10] * previous_sum[10] + previous_sum[11] * previous_sum[11] + previous_sum[12] * previous_sum[12] + previous_sum[13] * previous_sum[13] + previous_sum[14] * previous_sum[14] + previous_sum[15] * previous_sum[15] + previous_sum[16] * previous_sum[16] + previous_sum[17] * previous_sum[17] + previous_sum[18] * previous_sum[18]; previous_sum[6] = 0; previous_sum[5] = 0; previous_sum[4] = 0; previous_sum[3] = 0; previous_sum[2] = 0; previous_sum[1] = 0; previous_sum[0] = 0; } else if (n % 1000000 == 0) { previous_sum[6]++; sum_counter = previous_sum[6] + previous_sum[7] + previous_sum[8] + previous_sum[9] + previous_sum[10] + previous_sum[11] + previous_sum[12] + previous_sum[13] + previous_sum[14] + previous_sum[15] + previous_sum[16] + previous_sum[17] + previous_sum[18]; sum_square_counter = previous_sum[6] * previous_sum[6] + previous_sum[7] * previous_sum[7] + previous_sum[8] * previous_sum[8] + previous_sum[9] * previous_sum[9] + previous_sum[10] * previous_sum[10] + previous_sum[11] * previous_sum[11] + previous_sum[12] * previous_sum[12] + previous_sum[13] * previous_sum[13] + previous_sum[14] * previous_sum[14] + previous_sum[15] * previous_sum[15] + previous_sum[16] * previous_sum[16] + previous_sum[17] * previous_sum[17] + previous_sum[18] * previous_sum[18]; previous_sum[5] = 0; previous_sum[4] = 0; previous_sum[3] = 0; previous_sum[2] = 0; previous_sum[1] = 0; previous_sum[0] = 0; } else if (n % 100000 == 0) { previous_sum[5]++; sum_counter = previous_sum[5] + previous_sum[6] + previous_sum[7] + previous_sum[8] + previous_sum[9] + previous_sum[10] + previous_sum[11] + previous_sum[12] + previous_sum[13] + previous_sum[14] + previous_sum[15] + previous_sum[16] + previous_sum[17] + previous_sum[18]; sum_square_counter = previous_sum[5] * previous_sum[5] + previous_sum[6] * previous_sum[6] + previous_sum[7] * previous_sum[7] + previous_sum[8] * previous_sum[8] + previous_sum[9] * previous_sum[9] + previous_sum[10] * previous_sum[10] + previous_sum[11] * previous_sum[11] + previous_sum[12] * previous_sum[12] + previous_sum[13] * previous_sum[13] + previous_sum[14] * previous_sum[14] + previous_sum[15] * previous_sum[15] + previous_sum[16] * previous_sum[16] + previous_sum[17] * previous_sum[17] + previous_sum[18] * previous_sum[18]; previous_sum[4] = 0; previous_sum[3] = 0; previous_sum[2] = 0; previous_sum[1] = 0; previous_sum[0] = 0; } else if (n % 10000 == 0) { previous_sum[4]++; sum_counter = previous_sum[4] + previous_sum[5] + previous_sum[6] + previous_sum[7] + previous_sum[8] + previous_sum[9] + previous_sum[10] + previous_sum[11] + previous_sum[12] + previous_sum[13] + previous_sum[14] + previous_sum[15] + previous_sum[16] + previous_sum[17] + previous_sum[18]; sum_square_counter = previous_sum[4] * previous_sum[4] + previous_sum[5] * previous_sum[5] + previous_sum[6] * previous_sum[6] + previous_sum[7] * previous_sum[7] + previous_sum[8] * previous_sum[8] + previous_sum[9] * previous_sum[9] + previous_sum[12] * previous_sum[12] + previous_sum[13] * previous_sum[13] + previous_sum[14] * previous_sum[14] + previous_sum[15] * previous_sum[15] + previous_sum[16] * previous_sum[16] + previous_sum[17] * previous_sum[17] + previous_sum[18] * previous_sum[18]; previous_sum[3] = 0; previous_sum[2] = 0; previous_sum[1] = 0; previous_sum[0] = 0; } else if (n % 1000 == 0) { previous_sum[3]++; sum_counter = previous_sum[3] + previous_sum[4] + previous_sum[5] + previous_sum[6] + previous_sum[7] + previous_sum[8] + previous_sum[9] + previous_sum[10] + previous_sum[11] + previous_sum[12] + previous_sum[13] + previous_sum[14] + previous_sum[15] + previous_sum[16] + previous_sum[17] + previous_sum[18]; sum_square_counter = previous_sum[3] * previous_sum[3] + previous_sum[4] * previous_sum[4] + previous_sum[5] * previous_sum[5] + previous_sum[6] * previous_sum[6] + previous_sum[7] * previous_sum[7] + previous_sum[8] * previous_sum[8] + previous_sum[9] * previous_sum[9] + previous_sum[10] * previous_sum[10] + previous_sum[11] * previous_sum[11] + previous_sum[12] * previous_sum[12] + previous_sum[13] * previous_sum[13] + previous_sum[14] * previous_sum[14] + previous_sum[15] * previous_sum[15] + previous_sum[16] * previous_sum[16] + previous_sum[17] * previous_sum[17] + previous_sum[18] * previous_sum[18]; previous_sum[2] = 0; previous_sum[1] = 0; previous_sum[0] = 0; } else if (n % 100 == 0) { previous_sum[2]++; sum_counter = previous_sum[2] + previous_sum[3] + previous_sum[4] + previous_sum[5] + previous_sum[6] + previous_sum[7] + previous_sum[8] + previous_sum[9] + previous_sum[10] + previous_sum[11] + previous_sum[12] + previous_sum[13] + previous_sum[14] + previous_sum[15] + previous_sum[16] + previous_sum[17] + previous_sum[18]; sum_square_counter = previous_sum[2] * previous_sum[2] + previous_sum[3] * previous_sum[3] + previous_sum[4] * previous_sum[4] + previous_sum[5] * previous_sum[5] + previous_sum[6] * previous_sum[6] + previous_sum[7] * previous_sum[7] + previous_sum[8] * previous_sum[8] + previous_sum[9] * previous_sum[9] + previous_sum[10] * previous_sum[10] + previous_sum[11] * previous_sum[11] + previous_sum[12] * previous_sum[12] + previous_sum[13] * previous_sum[13] + previous_sum[14] * previous_sum[14] + previous_sum[15] * previous_sum[15] + previous_sum[16] * previous_sum[16] + previous_sum[17] * previous_sum[17] + previous_sum[18] * previous_sum[18]; previous_sum[1] = 0; previous_sum[0] = 0; } else if (n % 10 == 0) { previous_sum[1]++; sum_counter = previous_sum[1] + previous_sum[2] + previous_sum[3] + previous_sum[4] + previous_sum[5] + previous_sum[6] + previous_sum[7] + previous_sum[8] + previous_sum[9] + previous_sum[10] + previous_sum[11] + previous_sum[12] + previous_sum[13] + previous_sum[14] + previous_sum[15] + previous_sum[16] + previous_sum[17] + previous_sum[18]; sum_square_counter = previous_sum[1] * previous_sum[1] + previous_sum[2] * previous_sum[2] + previous_sum[3] * previous_sum[3] + previous_sum[4] * previous_sum[4] + previous_sum[5] * previous_sum[5] + previous_sum[6] * previous_sum[6] + previous_sum[7] * previous_sum[7] + previous_sum[8] * previous_sum[8] + previous_sum[9] * previous_sum[9] + previous_sum[10] * previous_sum[10] + previous_sum[11] * previous_sum[11] + previous_sum[12] * previous_sum[12] + previous_sum[13] * previous_sum[13] + previous_sum[14] * previous_sum[14] + previous_sum[15] * previous_sum[15] + previous_sum[16] * previous_sum[16] + previous_sum[17] * previous_sum[17] + previous_sum[18] * previous_sum[18]; previous_sum[0] = 0; } else { sum_counter++; sum_square_counter += ((n - 1) % 10) * 2 + 1; } // get_sum_and_sum_square_digits(n, sum, sum_square); // assert(sum == sum_counter && sum_square == sum_square_counter); if (prime_table[sum_counter] && prime_table[sum_square_counter]) { lucky_counter++; } } return lucky_counter; } void inout_lucky_numbers() { int n; cin >> n; long long a; long long b; while (n--) { cin >> a >> b; cout << count_lucky_number(a, b) << endl; } } int main() { inout_lucky_numbers(); return 0; }

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• Performance considerations for common SQL queries

  - by Jim Giercyk

  Originally posted on: http://geekswithblogs.net/NibblesAndBits/archive/2013/10/16/performance-considerations-for-common-sql-queries.aspxSQL offers many different methods to produce the same results.  There is a never-ending debate between SQL developers as to the “best way” or the “most efficient way” to render a result set.  Sometimes these disputes even come to blows….well, I am a lover, not a fighter, so I decided to collect some data that will prove which way is the best and most efficient.  For the queries below, I downloaded the test database from SQLSkills:  http://www.sqlskills.com/sql-server-resources/sql-server-demos/.  There isn’t a lot of data, but enough to prove my point: dbo.member has 10,000 records, and dbo.payment has 15,554.  Our result set contains 6,706 records. The following queries produce an identical result set; the result set contains aggregate payment information for each member who has made more than 1 payment from the dbo.payment table and the first and last name of the member from the dbo.member table.   /*************/ /* Sub Query  */ /*************/ SELECT  a.[Member Number] ,         m.lastname ,         m.firstname ,         a.[Number Of Payments] ,         a.[Average Payment] ,         a.[Total Paid] FROM    ( SELECT    member_no 'Member Number' ,                     AVG(payment_amt) 'Average Payment' ,                     SUM(payment_amt) 'Total Paid' ,                     COUNT(Payment_No) 'Number Of Payments'           FROM      dbo.payment           GROUP BY  member_no           HAVING    COUNT(Payment_No) > 1         ) a         JOIN dbo.member m ON a.[Member Number] = m.member_no         /***************/ /* Cross Apply  */ /***************/ SELECT  ca.[Member Number] ,         m.lastname ,         m.firstname ,         ca.[Number Of Payments] ,         ca.[Average Payment] ,         ca.[Total Paid] FROM    dbo.member m         CROSS APPLY ( SELECT    member_no 'Member Number' ,                                 AVG(payment_amt) 'Average Payment' ,                                 SUM(payment_amt) 'Total Paid' ,                                 COUNT(Payment_No) 'Number Of Payments'                       FROM      dbo.payment                       WHERE     member_no = m.member_no                       GROUP BY  member_no                       HAVING    COUNT(Payment_No) > 1                     ) ca /********/                    /* CTEs  */ /********/ ; WITH    Payments           AS ( SELECT   member_no 'Member Number' ,                         AVG(payment_amt) 'Average Payment' ,                         SUM(payment_amt) 'Total Paid' ,                         COUNT(Payment_No) 'Number Of Payments'                FROM     dbo.payment                GROUP BY member_no                HAVING   COUNT(Payment_No) > 1              ),         MemberInfo           AS ( SELECT   p.[Member Number] ,                         m.lastname ,                         m.firstname ,                         p.[Number Of Payments] ,                         p.[Average Payment] ,                         p.[Total Paid]                FROM     dbo.member m                         JOIN Payments p ON m.member_no = p.[Member Number]              )     SELECT  *     FROM    MemberInfo /************************/ /* SELECT with Grouping   */ /************************/ SELECT  p.member_no 'Member Number' ,         m.lastname ,         m.firstname ,         COUNT(Payment_No) 'Number Of Payments' ,         AVG(payment_amt) 'Average Payment' ,         SUM(payment_amt) 'Total Paid' FROM    dbo.payment p         JOIN dbo.member m ON m.member_no = p.member_no GROUP BY p.member_no ,         m.lastname ,         m.firstname HAVING  COUNT(Payment_No) > 1   We can see what is going on in SQL’s brain by looking at the execution plan.  The Execution Plan will demonstrate which steps and in what order SQL executes those steps, and what percentage of batch time each query takes.  SO….if I execute all 4 of these queries in a single batch, I will get an idea of the relative time SQL takes to execute them, and how it renders the Execution Plan.  We can settle this once and for all.  Here is what SQL did with these queries:   Not only did the queries take the same amount of time to execute, SQL generated the same Execution Plan for each of them.  Everybody is right…..I guess we can all finally go to lunch together!  But wait a second, I may not be a fighter, but I AM an instigator.     Let’s see how a table variable stacks up.  Here is the code I executed: /********************/ /*  Table Variable  */ /********************/ DECLARE @AggregateTable TABLE     (       member_no INT ,       AveragePayment MONEY ,       TotalPaid MONEY ,       NumberOfPayments MONEY     ) INSERT  @AggregateTable         SELECT  member_no 'Member Number' ,                 AVG(payment_amt) 'Average Payment' ,                 SUM(payment_amt) 'Total Paid' ,                 COUNT(Payment_No) 'Number Of Payments'         FROM    dbo.payment         GROUP BY member_no         HAVING  COUNT(Payment_No) > 1   SELECT  at.member_no 'Member Number' ,         m.lastname ,         m.firstname ,         at.NumberOfPayments 'Number Of Payments' ,         at.AveragePayment 'Average Payment' ,         at.TotalPaid 'Total Paid' FROM    @AggregateTable at         JOIN dbo.member m ON m.member_no = at.member_no In the interest of keeping things in groupings of 4, I removed the last query from the previous batch and added the table variable query.  Here’s what I got:     Since we first insert into the table variable, then we read from it, the Execution Plan renders 2 steps.  BUT, the combination of the 2 steps is only 22% of the batch.  It is actually faster than the other methods even though it is treated as 2 separate queries in the Execution Plan.  The argument I often hear against Table Variables is that SQL only estimates 1 row for the table size in the Execution Plan.  While this is true, the estimate does not come in to play until you read from the table variable.  In this case, the table variable had 6,706 rows, but it still outperformed the other queries.  People argue that table variables should only be used for hash or lookup tables.  The fact is, you have control of what you put IN to the variable, so as long as you keep it within reason, these results suggest that a table variable is a viable alternative to sub-queries. If anyone does volume testing on this theory, I would be interested in the results.  My suspicion is that there is a breaking point where efficiency goes down the tubes immediately, and it would be interesting to see where the threshold is. Coding SQL is a matter of style.  If you’ve been around since they introduced DB2, you were probably taught a little differently than a recent computer science graduate.  If you have a company standard, I strongly recommend you follow it.    If you do not have a standard, generally speaking, there is no right or wrong answer when talking about the efficiency of these types of queries, and certainly no hard-and-fast rule.  Volume and infrastructure will dictate a lot when it comes to performance, so your results may vary in your environment.  Download the database and try it!

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• Implementing custom "Remember Me" with Stripe

  - by Matt

  Implementing remember me with Stripe, while not using their Checkout (not supported on PhoneGap), seems to be fine using the path: First time: Request token on the client side using card info. Create customer on server side using token. Upon confirm, charge customer. Second time: Check if current user is Stripe customer by requesting the info from our server. If is Stripe customer, show "use credit card on file" instead of regular CC form. Upon confirm, charge customer. However, there is one important convenience items missing--last four digits of card number. Most sites inform you of the card you're using before making the payment, pretty important in case you have to switch out cards. I have seen that you can retrieve charges which would allow me to get the last four digits. Is it bad practice to pull that and display it? Are there alternative solutions anyone has in mind?

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• Project Euler 52: Ruby

  - by Ben Griswold

  In my attempt to learn Ruby out in the open, here’s my solution for Project Euler Problem 52.  Compared to Problem 51, this problem was a snap. Brute force and pretty quick… As always, any feedback is welcome. # Euler 52 # http://projecteuler.net/index.php?section=problems&id=52 # It can be seen that the number, 125874, and its double, # 251748, contain exactly the same digits, but in a # different order. # # Find the smallest positive integer, x, such that 2x, 3x, # 4x, 5x, and 6x, contain the same digits. timer_start = Time.now def contains_same_digits?(n) value = (n*2).to_s.split(//).uniq.sort.join 3.upto(6) do |i| return false if (n*i).to_s.split(//).uniq.sort.join != value end true end i = 100_000 answer = 0 while answer == 0 answer = i if contains_same_digits?(i) i+=1 end puts answer puts "Elapsed Time: #{(Time.now - timer_start)*1000} milliseconds"

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• Handwriting Recognition using Kernel Discriminant Analysis

  Demonstration of handwritten digit recognition using Kernel Discriminant Analysis and the Optical Recognition of Handwritten Digits Data Set from the UCI Machine Learning Repository.

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• Why are UUID / GUID's in the format they are?

  - by Xeoncross

  Globally Unique Identifiers (GUID) are a grouped string with a specific format which I assume has a security reason. A GUID is most commonly written in text as a sequence of hexadecimal digits separated into five groups, such as: 3F2504E0-4F89-11D3-9A0C-0305E82C3301 Why aren't GUID/UUID strings just random bytes encoded using hexadecimal of X length? This text notation contains the following fields, separated by hyphens: | Hex digits | Description |------------------------- | 8 | Data1 | 4 | Data2 | 4 | Data3 | 4 | Initial two bytes from Data4 | 12 | Remaining six bytes from Data4 There are also several versions of the UUID standards. Version 4 UUIDs are generally internally stored as a raw array of 128 bits, and typically displayed in a format something like: uuid:xxxxxxxx-xxxx-4xxx-yxxx-xxxxxxxxxxxx

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• How Red Hat Made Money in 2010

  CEO Jim Whitehurst speaks about how his Enterprise Linux company bucked the recession and grew revenues by double digits.

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• F# - Facebook Hacker Cup - Double Squares

  - by Jacob

  I'm working on strengthening my F#-fu and decided to tackle the Facebook Hacker Cup Double Squares problem. I'm having some problems with the run-time and was wondering if anyone could help me figure out why it is so much slower than my C# equivalent. There's a good description from another post; Source: Facebook Hacker Cup Qualification Round 2011 A double-square number is an integer X which can be expressed as the sum of two perfect squares. For example, 10 is a double-square because 10 = 3^2 + 1^2. Given X, how can we determine the number of ways in which it can be written as the sum of two squares? For example, 10 can only be written as 3^2 + 1^2 (we don't count 1^2 + 3^2 as being different). On the other hand, 25 can be written as 5^2 + 0^2 or as 4^2 + 3^2. You need to solve this problem for 0 = X = 2,147,483,647. Examples: 10 = 1 25 = 2 3 = 0 0 = 1 1 = 1 My basic strategy (which I'm open to critique on) is to; Create a dictionary (for memoize) of the input numbers initialzed to 0 Get the largest number (LN) and pass it to count/memo function Get the LN square root as int Calculate squares for all numbers 0 to LN and store in dict Sum squares for non repeat combinations of numbers from 0 to LN If sum is in memo dict, add 1 to memo Finally, output the counts of the original numbers. Here is the F# code (See code changes at bottom) I've written that I believe corresponds to this strategy (Runtime: ~8:10); open System open System.Collections.Generic open System.IO /// Get a sequence of values let rec range min max = seq { for num in [min .. max] do yield num } /// Get a sequence starting from 0 and going to max let rec zeroRange max = range 0 max /// Find the maximum number in a list with a starting accumulator (acc) let rec maxNum acc = function | [] -> acc | p::tail when p > acc -> maxNum p tail | p::tail -> maxNum acc tail /// A helper for finding max that sets the accumulator to 0 let rec findMax nums = maxNum 0 nums /// Build a collection of combinations; ie [1,2,3] = (1,1), (1,2), (1,3), (2,2), (2,3), (3,3) let rec combos range = seq { let count = ref 0 for inner in range do for outer in Seq.skip !count range do yield (inner, outer) count := !count + 1 } let rec squares nums = let dict = new Dictionary<int, int>() for s in nums do dict.[s] <- (s * s) dict /// Counts the number of possible double squares for a given number and keeps track of other counts that are provided in the memo dict. let rec countDoubleSquares (num: int) (memo: Dictionary<int, int>) = // The highest relevent square is the square root because it squared plus 0 squared is the top most possibility let maxSquare = System.Math.Sqrt((float)num) // Our relevant squares are 0 to the highest possible square; note the cast to int which shouldn't hurt. let relSquares = range 0 ((int)maxSquare) // calculate the squares up front; let calcSquares = squares relSquares // Build up our square combinations; ie [1,2,3] = (1,1), (1,2), (1,3), (2,2), (2,3), (3,3) for (sq1, sq2) in combos relSquares do let v = calcSquares.[sq1] + calcSquares.[sq2] // Memoize our relevant results if memo.ContainsKey(v) then memo.[v] <- memo.[v] + 1 // return our count for the num passed in memo.[num] // Read our numbers from file. //let lines = File.ReadAllLines("test2.txt") //let nums = [ for line in Seq.skip 1 lines -> Int32.Parse(line) ] // Optionally, read them from straight array let nums = [1740798996; 1257431873; 2147483643; 602519112; 858320077; 1048039120; 415485223; 874566596; 1022907856; 65; 421330820; 1041493518; 5; 1328649093; 1941554117; 4225; 2082925; 0; 1; 3] // Initialize our memoize dictionary let memo = new Dictionary<int, int>() for num in nums do memo.[num] <- 0 // Get the largest number in our set, all other numbers will be memoized along the way let maxN = findMax nums // Do the memoize let maxCount = countDoubleSquares maxN memo // Output our results. for num in nums do printfn "%i" memo.[num] // Have a little pause for when we debug let line = Console.Read() And here is my version in C# (Runtime: ~1:40: using System; using System.Collections.Generic; using System.Diagnostics; using System.IO; using System.Linq; using System.Text; namespace FBHack_DoubleSquares { public class TestInput { public int NumCases { get; set; } public List<int> Nums { get; set; } public TestInput() { Nums = new List<int>(); } public int MaxNum() { return Nums.Max(); } } class Program { static void Main(string[] args) { // Read input from file. //TestInput input = ReadTestInput("live.txt"); // As example, load straight. TestInput input = new TestInput { NumCases = 20, Nums = new List<int> { 1740798996, 1257431873, 2147483643, 602519112, 858320077, 1048039120, 415485223, 874566596, 1022907856, 65, 421330820, 1041493518, 5, 1328649093, 1941554117, 4225, 2082925, 0, 1, 3, } }; var maxNum = input.MaxNum(); Dictionary<int, int> memo = new Dictionary<int, int>(); foreach (var num in input.Nums) { if (!memo.ContainsKey(num)) memo.Add(num, 0); } DoMemoize(maxNum, memo); StringBuilder sb = new StringBuilder(); foreach (var num in input.Nums) { //Console.WriteLine(memo[num]); sb.AppendLine(memo[num].ToString()); } Console.Write(sb.ToString()); var blah = Console.Read(); //File.WriteAllText("out.txt", sb.ToString()); } private static int DoMemoize(int num, Dictionary<int, int> memo) { var highSquare = (int)Math.Floor(Math.Sqrt(num)); var squares = CreateSquareLookup(highSquare); var relSquares = squares.Keys.ToList(); Debug.WriteLine("Starting - " + num.ToString()); Debug.WriteLine("RelSquares.Count = {0}", relSquares.Count); int sum = 0; var index = 0; foreach (var square in relSquares) { foreach (var inner in relSquares.Skip(index)) { sum = squares[square] + squares[inner]; if (memo.ContainsKey(sum)) memo[sum]++; } index++; } if (memo.ContainsKey(num)) return memo[num]; return 0; } private static TestInput ReadTestInput(string fileName) { var lines = File.ReadAllLines(fileName); var input = new TestInput(); input.NumCases = int.Parse(lines[0]); foreach (var lin in lines.Skip(1)) { input.Nums.Add(int.Parse(lin)); } return input; } public static Dictionary<int, int> CreateSquareLookup(int maxNum) { var dict = new Dictionary<int, int>(); int square; foreach (var num in Enumerable.Range(0, maxNum)) { square = num * num; dict[num] = square; } return dict; } } } Thanks for taking a look. UPDATE Changing the combos function slightly will result in a pretty big performance boost (from 8 min to 3:45): /// Old and Busted... let rec combosOld range = seq { let rangeCache = Seq.cache range let count = ref 0 for inner in rangeCache do for outer in Seq.skip !count rangeCache do yield (inner, outer) count := !count + 1 } /// The New Hotness... let rec combos maxNum = seq { for i in 0..maxNum do for j in i..maxNum do yield i,j }

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• How to reference or vlookup a list of values based on a comma separated list of column references within a cell in excel?

  - by glallen

  I want to do a vlookup (or similar) against a column which is a list of values. This works fine for looking up a value from a single row, but I want to be able to look up multiple rows, sum the results, and divide by the number of rows referenced. For example: A B C D E F G [----given values----------------] [Work/Auth] [sum(vlookup(each(G),table,5)) /count(G)] [given vals] 1 Item Authorized OnHand Working Operational% DependencyOR% Dependencies 2 A 1 1 1 1 .55 B 3 B 10 5 5 .50 .55 C,D 4 C 100 75 50 .50 .60 D 5 D 10 10 6 .60 1 I want to be able to show an Operational Rate, and an operational rate of the systems each system depends on (F). In order to get a value for F, I want to sum over each value in column-E that was referenced by a dependency in column-G then divide by the number of dependencies in G. Column-G can have varying lengths, and will be a comma separated list of values from column-A. Is there any way to do this in excel?

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• Performing Aggregate Functions on Multi-Million Row Tables

  - by Daniel Short

  I'm having some serious performance issues with a multi-million row table that I feel I should be able to get results from fairly quick. Here's a run down of what I have, how I'm querying it, and how long it's taking: I'm running SQL Server 2008 Standard, so Partitioning isn't currently an option I'm attempting to aggregate all views for all inventory for a specific account over the last 30 days. All views are stored in the following table: CREATE TABLE [dbo].[LogInvSearches_Daily]( [ID] [bigint] IDENTITY(1,1) NOT NULL, [Inv_ID] [int] NOT NULL, [Site_ID] [int] NOT NULL, [LogCount] [int] NOT NULL, [LogDay] [smalldatetime] NOT NULL, CONSTRAINT [PK_LogInvSearches_Daily] PRIMARY KEY CLUSTERED ( [ID] ASC )WITH (PAD_INDEX = OFF, STATISTICS_NORECOMPUTE = OFF, IGNORE_DUP_KEY = OFF, ALLOW_ROW_LOCKS = ON, ALLOW_PAGE_LOCKS = ON, FILLFACTOR = 90) ON [PRIMARY] ) ON [PRIMARY] This table has 132,000,000 records, and is over 4 gigs. A sample of 10 rows from the table: ID Inv_ID Site_ID LogCount LogDay -------------------- ----------- ----------- ----------- ----------------------- 1 486752 48 14 2009-07-21 00:00:00 2 119314 51 16 2009-07-21 00:00:00 3 313678 48 25 2009-07-21 00:00:00 4 298863 0 1 2009-07-21 00:00:00 5 119996 0 2 2009-07-21 00:00:00 6 463777 534 7 2009-07-21 00:00:00 7 339976 503 2 2009-07-21 00:00:00 8 333501 570 4 2009-07-21 00:00:00 9 453955 0 12 2009-07-21 00:00:00 10 443291 0 4 2009-07-21 00:00:00 (10 row(s) affected) I have the following index on LogInvSearches_Daily: /****** Object: Index [IX_LogInvSearches_Daily_LogDay] Script Date: 05/12/2010 11:08:22 ******/ CREATE NONCLUSTERED INDEX [IX_LogInvSearches_Daily_LogDay] ON [dbo].[LogInvSearches_Daily] ( [LogDay] ASC ) INCLUDE ( [Inv_ID], [LogCount]) WITH (PAD_INDEX = OFF, STATISTICS_NORECOMPUTE = OFF, SORT_IN_TEMPDB = OFF, IGNORE_DUP_KEY = OFF, DROP_EXISTING = OFF, ONLINE = OFF, ALLOW_ROW_LOCKS = ON, ALLOW_PAGE_LOCKS = ON) ON [PRIMARY] I need to pull inventory only from the Inventory for a specific account id. I have an index on the Inventory as well. I'm using the following query to aggregate the data and give me the top 5 records. This query is currently taking 24 seconds to return the 5 rows: StmtText ---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- SELECT TOP 5 Sum(LogCount) AS Views , DENSE_RANK() OVER(ORDER BY Sum(LogCount) DESC, Inv_ID DESC) AS Rank , Inv_ID FROM LogInvSearches_Daily D (NOLOCK) WHERE LogDay DateAdd(d, -30, getdate()) AND EXISTS( SELECT NULL FROM propertyControlCenter.dbo.Inventory (NOLOCK) WHERE Acct_ID = 18731 AND Inv_ID = D.Inv_ID ) GROUP BY Inv_ID (1 row(s) affected) StmtText ---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- |--Top(TOP EXPRESSION:((5))) |--Sequence Project(DEFINE:([Expr1007]=dense_rank)) |--Segment |--Segment |--Sort(ORDER BY:([Expr1006] DESC, [D].[Inv_ID] DESC)) |--Stream Aggregate(GROUP BY:([D].[Inv_ID]) DEFINE:([Expr1006]=SUM([LOALogs].[dbo].[LogInvSearches_Daily].[LogCount] as [D].[LogCount]))) |--Sort(ORDER BY:([D].[Inv_ID] ASC)) |--Nested Loops(Inner Join, OUTER REFERENCES:([D].[Inv_ID])) |--Nested Loops(Inner Join, OUTER REFERENCES:([Expr1011], [Expr1012], [Expr1010])) | |--Compute Scalar(DEFINE:(([Expr1011],[Expr1012],[Expr1010])=GetRangeWithMismatchedTypes(dateadd(day,(-30),getdate()),NULL,(6)))) | | |--Constant Scan | |--Index Seek(OBJECT:([LOALogs].[dbo].[LogInvSearches_Daily].[IX_LogInvSearches_Daily_LogDay] AS [D]), SEEK:([D].[LogDay] > [Expr1011] AND [D].[LogDay] < [Expr1012]) ORDERED FORWARD) |--Index Seek(OBJECT:([propertyControlCenter].[dbo].[Inventory].[IX_Inventory_Acct_ID]), SEEK:([propertyControlCenter].[dbo].[Inventory].[Acct_ID]=(18731) AND [propertyControlCenter].[dbo].[Inventory].[Inv_ID]=[LOA (13 row(s) affected) I tried using a CTE to pick up the rows first and aggregate them, but that didn't run any faster, and gives me essentially the same execution plan. (1 row(s) affected) StmtText ---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- --SET SHOWPLAN_TEXT ON; WITH getSearches AS ( SELECT LogCount -- , DENSE_RANK() OVER(ORDER BY Sum(LogCount) DESC, Inv_ID DESC) AS Rank , D.Inv_ID FROM LogInvSearches_Daily D (NOLOCK) INNER JOIN propertyControlCenter.dbo.Inventory I (NOLOCK) ON Acct_ID = 18731 AND I.Inv_ID = D.Inv_ID WHERE LogDay DateAdd(d, -30, getdate()) -- GROUP BY Inv_ID ) SELECT Sum(LogCount) AS Views, Inv_ID FROM getSearches GROUP BY Inv_ID (1 row(s) affected) StmtText ------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------ |--Stream Aggregate(GROUP BY:([D].[Inv_ID]) DEFINE:([Expr1004]=SUM([LOALogs].[dbo].[LogInvSearches_Daily].[LogCount] as [D].[LogCount]))) |--Sort(ORDER BY:([D].[Inv_ID] ASC)) |--Nested Loops(Inner Join, OUTER REFERENCES:([D].[Inv_ID])) |--Nested Loops(Inner Join, OUTER REFERENCES:([Expr1008], [Expr1009], [Expr1007])) | |--Compute Scalar(DEFINE:(([Expr1008],[Expr1009],[Expr1007])=GetRangeWithMismatchedTypes(dateadd(day,(-30),getdate()),NULL,(6)))) | | |--Constant Scan | |--Index Seek(OBJECT:([LOALogs].[dbo].[LogInvSearches_Daily].[IX_LogInvSearches_Daily_LogDay] AS [D]), SEEK:([D].[LogDay] > [Expr1008] AND [D].[LogDay] < [Expr1009]) ORDERED FORWARD) |--Index Seek(OBJECT:([propertyControlCenter].[dbo].[Inventory].[IX_Inventory_Acct_ID] AS [I]), SEEK:([I].[Acct_ID]=(18731) AND [I].[Inv_ID]=[LOALogs].[dbo].[LogInvSearches_Daily].[Inv_ID] as [D].[Inv_ID]) ORDERED FORWARD) (8 row(s) affected) (1 row(s) affected) So given that I'm getting good Index Seeks in my execution plan, what can I do to get this running faster? Thanks, Dan

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• Supporting Piping (A Useful Hello World)

  - by blastthisinferno

  I am trying to write a collection of simple C++ programs that follow the basic Unix philosophy by: Make each program do one thing well. Expect the output of every program to become the input to another, as yet unknown, program. I'm having an issue trying to get the output of one to be the input of the other, and getting the output of one be the input of a separate instance of itself. Very briefly, I have a program add which takes arguments and spits out the summation. I want to be able to pipe the output to another add instance. ./add 1 2 | ./add 3 4 That should yield 6 but currently yields 10. I've encountered two problems: The cin waits for user input from the console. I don't want this, and haven't been able to find a simple example showing a the use of standard input stream without querying the user in the console. If someone knows of an example please let me know. I can't figure out how to use standard input while supporting piping. Currently, it appears it does not work. If I issue the command ./add 1 2 | ./add 3 4 it results in 7. The relevant code is below: add.cpp snippet // ... COMMAND LINE PROCESSING ... std::vector<double> numbers = multi.getValue(); // using TCLAP for command line parsing if (numbers.size() > 0) { double sum = numbers[0]; double arg; for (int i=1; i < numbers.size(); i++) { arg = numbers[i]; sum += arg; } std::cout << sum << std::endl; } else { double input; // right now this is test code while I try and get standard input streaming working as expected while (std::cin) { std::cin >> input; std::cout << input << std::endl; } } // ... MORE IRRELEVANT CODE ... So, I guess my question(s) is does anyone see what is incorrect with this code in order to support piping standard input? Are there some well known (or hidden) resources that explain clearly how to implement an example application supporting the basic Unix philosophy? @Chris Lutz I've changed the code to what's below. The problem where cin still waits for user input on the console, and doesn't just take from the standard input passed from the pipe. Am I missing something trivial for handling this? I haven't tried Greg Hewgill's answer yet, but don't see how that would help since the issue is still with cin. // ... COMMAND LINE PROCESSING ... std::vector<double> numbers = multi.getValue(); // using TCLAP for command line parsing double sum = numbers[0]; double arg; for (int i=1; i < numbers.size(); i++) { arg = numbers[i]; sum += arg; } // right now this is test code while I try and get standard input streaming working as expected while (std::cin) { std::cin >> arg; std::cout << arg << std::endl; } std::cout << sum << std::endl; // ... MORE IRRELEVANT CODE ...

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• LINQ and Aggregate function

  - by vik20000in

  LINQ also provides with itself important aggregate function. Aggregate function are function that are applied over a sequence like and return only one value like Average, count, sum, Maximum etc…Below are some of the Aggregate functions provided with LINQ and example of their implementation. Count     int[] primeFactorsOf300 = { 2, 2, 3, 5, 5 };     int uniqueFactors = primeFactorsOf300.Distinct().Count();The below example provided count for only odd number.     int[] primeFactorsOf300 = { 2, 2, 3, 5, 5 };     int uniqueFactors = primeFactorsOf300.Distinct().Count(n => n%2 = 1);  Sum     int[] numbers = { 5, 4, 1, 3, 9, 8, 6, 7, 2, 0 };        double numSum = numbers.Sum();  Minimum      int minNum = numbers.Min(); Maximum      int maxNum = numbers.Max();Average      double averageNum = numbers.Average();  Aggregate      double[] doubles = { 1.7, 2.3, 1.9, 4.1, 2.9 };     double product = doubles.Aggregate((runningProduct, nextFactor) => runningProduct * nextFactor);  Vikram

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• BizTalk 2009 - Scoped Record Counting in Maps

  - by StuartBrierley

  Within BizTalk there is a functoid called Record Count that will return the number of instances of a repeated record or repeated element that occur in a message instance. The input to this functoid is the record or element to be counted. As an example take the following Source schema, where the Source message has a repeated record called Box and each Box has a repeated element called Item: An instance of this Source schema may look as follows; 2 box records - one with 2 items and one with only 1 item. Our destination schema has a number of elements and a repeated box record.  The top level elements contain totals for the number of boxes and the overall number of items.  Each box record contains a single element representing the number of items in that box. Using the Record Count functoid it is easy to map the top level elements, producing the expected totals of 2 boxes and 3 items: We now need to map the total number of items per box, but how will we do this?  We have already seen that the record count functoid returns the total number of instances for the entire message, and unfortunately it does not allow you to specify a scoping parameter.  In order to acheive Scoped Record Counting we will need to make use of a combination of functoids. As you can see above, by linking to a Logical Existence functoid from the record/element to be counted we can then feed the output into a Value Mapping functoid.  Set the other Value Mapping parameter to "1" and link the output to a Cumulative Sum functoid. Set the other Cumulative Sum functoid parameter to "1" to limit the scope of the Cumulative Sum. This gives us the expected results of Items per Box of 2 and 1 respectively. I ran into this issue with a larger schema on a more complex map, but the eventual solution is still the same.  Hopefully this simplified example will act as a good reminder to me and save someone out there a few minutes of brain scratching.

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• how to architect this to make it unit testable

  - by SOfanatic

  I'm currently working on a project where I'm receiving an object via web service (WSDL). The overall process is the following: Receive object - add/delete/update parts (or all) of it - and return the object with the changes made. The thing is that sometimes these changes are complicated and there is some logic involved, other databases, other web services, etc. so to facilitate this I'm creating a custom object that mimics the original one but has some enhanced functionality to make some things easier. So I'm trying to have this process: Receive original object - convert/copy it to custom object - add/delete/update - convert/copy it back to original object - return original object. Example: public class Row { public List<Field> Fields { get; set; } public string RowId { get; set; } public Row() { this.Fields = new List<Field>(); } } public class Field { public string Number { get; set; } public string Value { get; set; } } So for example, one of the "actions" to perform on this would be to find all Fields in a Row that match a Value equal to something, and update them with some other value. I have a CustomRow class that represents the Row class, how can I make this class unit testable? Do I have to create an interface ICustomRow to mock it in the unit test? If one of the actions is to sum all of the Values in the Fields that have a Number equal to 10, like this function, how can design the custom class to facilitate unit tests. Sample function: public int Sum(FieldNumber number) { return row.Fields.Where(x => x.FieldNumber.Equals(number)).Sum(x => x.FieldValue); } Am I approaching this the wrong way?

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• how to organize rendering

  - by Irbis

  I use a deferred rendering. During g-buffer stage my rendering loop for a sponza model (obj format) looks like this: int i = 0; int sum = 0; map<string, mtlItem *>::const_iterator itrEnd = mtl.getIteratorEnd(); for(map<string, mtlItem *>::const_iterator itr = mtl.getIteratorBegin(); itr != itrEnd; ++itr) { glActiveTexture(GL_TEXTURE0 + 0); glBindTexture(GL_TEXTURE_2D, itr->second->map_KdId); glDrawElements(GL_TRIANGLES, indicesCount[i], GL_UNSIGNED_INT, (GLvoid*)(sum * 4)); sum += indicesCount[i]; ++i; glBindTexture(GL_TEXTURE_2D, 0); } I sorted faces based on materials. I switch only a diffuse texture but I can place there more material properties. Is it a good approach ? I also wonder how to handle a different kind of materials, for example: some material use a normal map, other doesn't use. Should I have a different shaders for them ?

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• Paypal date formatting: what is "HH:MM:SS DD Mmm YY, YYYY PST"?

  - by MTsoul

  I'm looking at the Paypal IPN docs, and it says the datetime stamps of their strings are formatted as: HH:MM:SS DD Mmm YY, YYYY PST So year is specified twice? Once in double digits, and another with 4 digits? This looks bizarre.

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• How to update multiple rows with different values in mysql?

  - by user210481

  I have a table with a column 'A'. Some rows have 14 digits for the column 'A' and some have only 12. I need to transform all the entries to 14 digits. The datatype is varchar I would like to update all the rows at once (one query), adding zeros before the first digit, so an entry like 012345678910 would become 00012345678910. Is it possible to do it in one single query? Thanks

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• Ab Initio - Formatting a number in Left alignment

  - by Veera

  I have a requirement in Ab Initio to format a number in left alignment. I shouldn't be using String conversion (as Strings are left aligned by default), as it might cause compatibility problems in the other end. For example, if my Field has 7 bytes length, and I'm getting only two digits as my input, then these two digits should go into the first two bytes of my field (left aligned), instead of the last two bytes. So, is there any in-built function in Ab Initio, that can format a number as left aligned?

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• Exception in thread "main" java.lang.StackOverflowError

  - by Ray.R.Chua

  I have a piece of code and I could not figure out why it is giving me Exception in thread "main" java.lang.StackOverflowError. This is the question: Given a positive integer n, prints out the sum of the lengths of the Syracuse sequence starting in the range of 1 to n inclusive. So, for example, the call: lengths(3) will return the the combined length of the sequences: 1 2 1 3 10 5 16 8 4 2 1 which is the value: 11. lengths must throw an IllegalArgumentException if its input value is less than one. My Code: import java.util.HashMap; public class Test { HashMap<Integer,Integer> syraSumHashTable = new HashMap<Integer,Integer>(); public Test(){ } public int lengths(int n)throws IllegalArgumentException{ int sum =0; if(n < 1){ throw new IllegalArgumentException("Error!! Invalid Input!"); } else{ for(int i =1; i<=n;i++){ if(syraSumHashTable.get(i)==null) { syraSumHashTable.put(i, printSyra(i,1)); sum += (Integer)syraSumHashTable.get(i); } else{ sum += (Integer)syraSumHashTable.get(i); } } return sum; } } private int printSyra(int num, int count){ int n = num; if(n == 1){ return count; } else{ if(n%2==0){ return printSyra(n/2, ++count); } else{ return printSyra((n*3)+1, ++count) ; } } } } Driver code: public static void main(String[] args) { // TODO Auto-generated method stub Test s1 = new Test(); System.out.println(s1.lengths(90090249)); //System.out.println(s1.lengths(5)); } . I know the problem lies with the recursion. The error does not occur if the input is a small value, example: 5. But when the number is huge, like 90090249, I got the Exception in thread "main" java.lang.StackOverflowError. Thanks all for your help. :) I almost forgot the error msg: Exception in thread "main" java.lang.StackOverflowError at Test.printSyra(Test.java:60) at Test.printSyra(Test.java:65) at Test.printSyra(Test.java:60) at Test.printSyra(Test.java:65) at Test.printSyra(Test.java:60) at Test.printSyra(Test.java:60) at Test.printSyra(Test.java:60) at Test.printSyra(Test.java:60)

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• Calling a function within a class

  - by JM4

  I am having trouble calling a specific function within a class. The call is made: case "Mod10": if (!validateCreditCard($fields[$field_name])) $errors[] = $error_message; break; and the class code is: class CreditCardValidationSolution { var $CCVSNumber = ''; var $CCVSNumberLeft = ''; var $CCVSNumberRight = ''; var $CCVSType = ''; var $CCVSError = ''; function validateCreditCard($Number) { $this->CCVSNumber = ''; $this->CCVSNumberLeft = ''; $this->CCVSNumberRight = ''; $this->CCVSType = ''; $this->CCVSError = ''; // Catch malformed input. if (empty($Number) || !is_string($Number)) { $this->CCVSError = $CCVSErrNumberString; return FALSE; } // Ensure number doesn't overrun. $Number = substr($Number, 0, 20); // Remove non-numeric characters. $this->CCVSNumber = preg_replace('/[^0-9]/', '', $Number); // Set up variables. $this->CCVSNumberLeft = substr($this->CCVSNumber, 0, 4); $this->CCVSNumberRight = substr($this->CCVSNumber, -4); $NumberLength = strlen($this->CCVSNumber); $DoChecksum = 'Y'; // Mod10 checksum process... if ($DoChecksum == 'Y') { $Checksum = 0; // Add even digits in even length strings or odd digits in odd length strings. for ($Location = 1 - ($NumberLength % 2); $Location < $NumberLength; $Location += 2) { $Checksum += substr($this->CCVSNumber, $Location, 1); } // Analyze odd digits in even length strings or even digits in odd length strings. for ($Location = ($NumberLength % 2); $Location < $NumberLength; $Location += 2) { $Digit = substr($this->CCVSNumber, $Location, 1) * 2; if ($Digit < 10) { $Checksum += $Digit; } else { $Checksum += $Digit - 9; } } // Checksums not divisible by 10 are bad. if ($Checksum % 10 != 0) { $this->CCVSError = $CCVSErrChecksum; return FALSE; } } return TRUE; } } When I run the application - I get the following message: Fatal error: Call to undefined function validateCreditCard() in C:\xampp\htdocs\validation.php on line 339 any ideas?

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• Django: FloatField or DecimalFied for Currency ?

  - by Hellnar

  I am curious which one would be better fitting as a currency field ? I will do simple operations such as taking difference, the percentage between old and new prices. I plan to keep two digits after the zero (ie 10.50) and majority of the time if these digits are zero, I will be hiding these numbers and display it as "10"

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• In SQL server, to convert a varchar which have this format (nnn:nn:nn)

  - by user1688917

  I have this varchar format as time accumulation and i want to convert it to an integer to do a SUM and get the total time for a group. The fist part which may be 1, 2, 3, 4 or even five digits represent the accumulation of Hours and then seperated by a colon. then come the second part which is accumulation of minutes and last accumulation of seconds (2 digits each). How to convert this to integer in one query if possile.

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• Ruby on Rails: How do you add add zeros in front of a number if it's under 10?

  - by sjsc

  I'm looking to convert single digit numbers to two-digit numbers like so: 9 ==> 09 5 ==> 05 12 == 12 4 ==> 04 I figure I could put a bunch of if-else statements (if number is under 10, then do a gsub) but figure that's horrible coding. I know Rails has number_with_precision but I see that it only applies to decimal numbers. Any ideas on how to convert single-digits to two-digits?

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