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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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• Find kth smallest element in a binary search tree in Optimum way

  - by Bragaadeesh

  Hi, I need to find the kth smallest element in the binary search tree without using any static/global variable. How to achieve it efficiently? The solution that I have in my mind is doing the operation in O(n), the worst case since I am planning to do an inorder traversal of the entire tree. But deep down I feel that I am not using the BST property here. Is my assumptive solution correct or is there a better one available ?

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• Write a function that compares two strings and returns a third string containing only the letters th

  - by Pritam

  Hi All, I got this homework. And have solved it in following way. I need your comments whether it is a good approach or I need to use any other data sturcture to solve it in better way. public string ReturnCommon(string firstString, string scndString) { StringBuilder newStb = new StringBuilder(); if (firstString != null && scndString != null) { foreach (char ichar in firstString) { if (!newStb.ToString().Contains(ichar) && scndString.Contains(ichar)) newStb.Append(ichar); } } return newStb.ToString(); }

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• Point in polygon OR point on polygon using LINQ

  - by wageoghe

  As noted in an earlier question, How to Zip enumerable with itself, I am working on some math algorithms based on lists of points. I am currently working on point in polygon. I have the code for how to do that and have found several good references here on SO, such as this link Hit test. So, I can figure out whether or not a point is in a polygon. As part of determining that, I want to determine if the point is actually on the polygon. This I can also do. If I can do all of that, what is my question you might ask? Can I do it efficiently using LINQ? I can already do something like the following (assuming a Pairwise extension method as described in my earlier question as well as in links to which my question/answers links, and assuming a Position type that has X and Y members). I have not tested much, so the lambda might not be 100% correct. Also, it does not take very small differences into account. public static PointInPolygonLocation PointInPolygon(IEnumerable<Position> pts, Position pt) { int numIntersections = pts.Pairwise( (p1, p2) => { if (p1.Y != p2.Y) { if ((p1.Y >= pt.Y && p2.Y < pt.Y) || (p1.Y < pt.Y && p2.Y >= pt.Y)) { if (p1.X < p1.X && p2.X < pt.X) { return 1; } if (p1.X < pt.X || p2.X < pt.X) { if (((pt.Y - p1.Y) * ((p1.X - p2.X) / (p1.Y - p2.Y)) * p1.X) < pt.X) { return 1; } } } } return 0; }).Sum(); if (numIntersections % 2 == 0) { return PointInPolygonLocation.Outside; } else { return PointInPolygonLocation.Inside; } } This function, PointInPolygon, takes the input Position, pt, iterates over the input sequence of position values, and uses the Jordan Curve method to determine how many times a ray extended from pt to the left intersects the polygon. The lambda expression will yield, into the "zipped" list, 1 for every segment that is crossed, and 0 for the rest. The sum of these values determines if pt is inside or outside of the polygon (odd == inside, even == outside). So far, so good. Now, for any consecutive pairs of position values in the sequence (i.e. in any execution of the lambda), we can also determine if pt is ON the segment p1, p2. If that is the case, we can stop the calculation because we have our answer. Ultimately, my question is this: Can I perform this calculation (maybe using Aggregate?) such that we will only iterate over the sequence no more than 1 time AND can we stop the iteration if we encounter a segment that pt is ON? In other words, if pt is ON the very first segment, there is no need to examine the rest of the segments because we have the answer. It might very well be that this operation (particularly the requirement/desire to possibly stop the iteration early) does not really lend itself well to the LINQ approach. It just occurred to me that maybe the lambda expression could yield a tuple, the intersection value (1 or 0 or maybe true or false) and the "on" value (true or false). Maybe then I could use TakeWhile(anontype.PointOnPolygon == false). If I Sum the tuples and if ON == 1, then the point is ON the polygon. Otherwise, the oddness or evenness of the sum of the other part of the tuple tells if the point is inside or outside.

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• Best practice to calculate the average speed from GPS coordinates

  - by Sebi

  i have here a device which can give me gps coordinates. the time intervall i can define. i want to use it to calculate the average speed during driving or travelling by car. actually i used a orthodrome formula to calculate the distance between two points and then divided it by the given time intervall. by the implemenation i followed this term (http://de.wikipedia.org/wiki/Orthodrome#Genauere_Formel_zur_Abstandsberechnung_auf_der_Erde). Unfortunately i could only find a german link, but i think the formula should be understandable in any language ;) Unfortunately, using this formula and a time intverall of 1 seconds gives very unprecises results. the speed while walking is between 1 km/h and 20km/h. So i wonder if there is a general reference how to implement distance calculation between two gps coordinates (i found something similar on SO) and particulary, which is the best time intervall to update the GPS coordiantes.

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• How do I detect overflow while multiplying two 2's complement integers?

  - by eSKay

  I want to multiply two numbers, and detect if there was an overflow. What is the simplest way to do that?

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• Dynamic programming - Coin change decision problem?

  - by Tony

  I'm reviewing some old notes from my algorithms course and the dynamic programming problems are seeming a bit tricky to me. I have a problem where we have an unlimited supply of coins, with some denominations x1, x2, ... xn and we want to make change for some value X. We are trying to design a dynamic program to decide whether change for X can be made or not (not minimizing the number of coins, or returning which coins, just true or false). I've done some thinking about this problem, and I can see a recursive method of doing this where it's something like... MakeChange(X, x[1..n this is the coins]) for (int i = 1; i < n; i++) { if ( (X - x[i] ==0) || MakeChange(X - x[i]) ) return true; } return false; Converting this a dynamic program is not coming so easily to me. How might I approach this?

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• Finding anagaram(s) of dictionary words

  - by Codenotguru

  How can I take an input word (or sequence of letters) and output a word from a dictionary that contains exactly those letters? Does java has an English dictionary class (list of words) that I can use, or are there open source implementations of this? How can I optimize my code if this needs to be done repeatedly?

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• Find the "largest" dense sub matrix in a large sparse matrix

  - by BCS

  Given a large sparse matrix (say 10k+ by 1M+) I need to find a subset, not necessarily continuous, of the rows and columns that form a dense matrix (all non-zero elements). I want this sub matrix to be as large as possible (not the largest sum, but the largest number of elements) within some aspect ratio constraints. Are there any known exact or aproxamate solutions to this problem? A quick scan on Google seems to give a lot of close-but-not-exactly results. What terms should I be looking for? edit: Just to clarify; the sub matrix need not be continuous. In fact the row and column order is completely arbitrary so adjacency is completely irrelevant. A thought based on Chad Okere's idea Order the rows from largest count to smallest count (not necessary but might help perf) Select two rows that have a "large" overlap Add all other rows that won't reduce the overlap Record that set Add whatever row reduces the overlap by the least Repeat at #3 until the result gets to small Start over at #2 with a different starting pair Continue until you decide the result is good enough

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• How to Zip one IEnumerable with itself

  - by wageoghe

  I am implementing some math algorithms based on lists of points, like Distance, Area, Centroid, etc. Just like in this post: http://stackoverflow.com/questions/2227828/find-the-distance-required-to-navigate-a-list-of-points-using-linq That post describes how to calculate the total distance of a sequence of points (taken in order) by essentially zipping the sequence "with itself", generating the sequence for Zip by offsetting the start position of the original IEnumerable by 1. So, given the Zip extension in .Net 4.0, assuming Point for the point type, and a reasonable Distance formula, you can make calls like this to generate a sequence of distances from one point to the next and then to sum the distances: var distances = points.Zip(points.Skip(1),Distance); double totalDistance = distances.Sum(); Area and Centroid calculations are similar in that they need to iterate over the sequence, processing each pair of points (points[i] and points[i+1]). I thought of making a generic IEnumerable extension suitable for implementing these (and possibly other) algorithms that operate over sequences, taking two items at a time (points[0] and points[1], points[1] and points[2], ..., points[n-1] and points[n] (or is it n-2 and n-1 ...) and applying a function. My generic iterator would have a similar signature to Zip, but it would not receive a second sequence to zip with as it is really just going to zip with itself. My first try looks like this: public static IEnumerable<TResult> ZipMyself<TSequence, TResult>(this IEnumerable<TSequence> seq, Func<TSequence, TSequence, TResult> resultSelector) { return seq.Zip(seq.Skip(1),resultSelector); } With my generic iterator in place, I can write functions like this: public static double Length(this IEnumerable<Point> points) { return points.ZipMyself(Distance).Sum(); } and call it like this: double d = points.Length(); and double GreensTheorem(Point p1, Point p1) { return p1.X * p2.Y - p1.Y * p2.X; } public static double SignedArea(this IEnumerable<Point> points) { return points.ZipMyself(GreensTheorem).Sum() / 2.0 } public static double Area(this IEnumerable<Point> points) { return Math.Abs(points.SignedArea()); } public static bool IsClockwise(this IEnumerable<Point> points) { return SignedArea(points) < 0; } and call them like this: double a = points.Area(); bool isClockwise = points.IsClockwise(); In this case, is there any reason NOT to implement "ZipMyself" in terms of Zip and Skip(1)? Is there already something in LINQ that automates this (zipping a list with itself) - not that it needs to be made that much easier ;-) Also, is there better name for the extension that might reflect that it is a well-known pattern (if, indeed it is a well-known pattern)? Had a link here for a StackOverflow question about area calculation. It is question 2432428. Also had a link to Wikipedia article on Centroid. Just go to Wikipedia and search for Centroid if interested. Just starting out, so don't have enough rep to post more than one link,

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• Looking for more details about "Group varint encoding/decoding" presented in Jeff's slides

  - by Mickey Shine

  I noticed that in Jeff's slides "Challenges in Building Large-Scale Information Retrieval Systems", which can also be downloaded here: http://research.google.com/people/jeff/WSDM09-keynote.pdf, a method of integers compression called "group varint encoding" was mentioned. It was said much faster than 7 bits per byte integer encoding (2X more). I am very interested in this and looking for an implementation of this, or any more details that could help me implement this by myself. I am not a pro and new to this, and any help is welcome!

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• Find largest rectangle containing all zero's in an N X N binary matrix

  - by Rajendra

  Given an N X N binary matrix (containing only 0's or 1's). How can we go about finding largest rectangle containing all 0's? Example: I 0 0 0 0 1 0 0 0 1 0 0 1 II->0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 <--IV 0 0 1 0 0 0 IV is a 6 X 6 binary matrix, return value in this case will be Cell 1: (2, 1) and Cell 2: (4, 4). Resulting sub-matrix can be square or rectangle. Return value can be size of the largest sub-matrix of all 0's also, for example, here 3 X 4.

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• commands&creating pointer [closed]

  - by gcc

  input 23 3 4 4 42 n 23 0 9 9 n n n 3 9 9 x //according to input,i should create int pointer arrays. pointer arrays // starting from 1 (that is initial arrays is arrays[1].when program sees n ,it // must be jumb to arrays 2 // the first int input 23 is num_arrays which used in malloc(sizeof(int)*num_arrays expected output: elements of arrays[1] 3 4 5 42 elements of arrays[2] 23 0 9 9 elements of arrays[5] 3 9 9 another input 12 2 3 4 n n 2 3 4 n 12 3 x expected output elements of arrays[1] 2 3 4 elements of arrays[3] 2 3 4 elements of arrays[4] 12 3 specification: x is stopper n is comman to create new pointer array i am new in this site anyone help me how can i write

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• How is counting sort a stable sort?

  - by eSKay

  Suppose my input is (a,b and c to distinguish between equal keys) 1 6a 8 3 6b 0 6c 4 My counting sort will save as (discarding the a,b and c info!!) 0(1) 1(1) 3(1) 4(1) 6(3) 8(1) which will give me the result 0 1 3 4 6 6 6 8 So, how is this stable sort? I am not sure how it is "maintaining the relative order of records with equal keys." Please explain.

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• Sha or Md5 algorithm i need to encrypt and decrypt in flex

  - by praveen

  Hi I am developing my application in flex and JSP, so when I am passing values through HTTP Service Post method with request object but these values are tracing and modifying by testing team so I am planning to encrypt values in flex and decrypt it in jsp.so is there any algorithms like SHA or MD5 more secure algorithms, so please send any code or related links it is very useful to me. I am using like httpService = new HTTPService; httpService.request = new Object; httpService.request.task = "doInvite"; httpService.request.email = emailInput.text; httpService.request.firstName = firstNameInput.text; httpService.request.lastName = lastNameInput.text; httpService.send(); So is there any other way to give more secure ,please help me in this,Thanks in Advance.

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• C# average function for large numbers without overflow exception

  - by Ron Klein

  .NET Framework 3.5. I'm trying to calculate the average of some pretty large numbers. For instance: using System; using System.Linq; class Program { static void Main(string[] args) { var items = new long[] { long.MaxValue - 100, long.MaxValue - 200, long.MaxValue - 300 }; try { var avg = items.Average(); Console.WriteLine(avg); } catch (OverflowException ex) { Console.WriteLine("can't calculate that!"); } Console.ReadLine(); } } Obviously, the mathematical result is 9223372036854775607 (long.MaxValue - 200), but I get an exception there. This is because the implementation (on my machine) to the Average extension method, as inspected by .NET Reflector is: public static double Average(this IEnumerable<long> source) { if (source == null) { throw Error.ArgumentNull("source"); } long num = 0L; long num2 = 0L; foreach (long num3 in source) { num += num3; num2 += 1L; } if (num2 <= 0L) { throw Error.NoElements(); } return (((double) num) / ((double) num2)); } I know I can use a BigInt library (yes, I know that it is included in .NET Framework 4.0, but I'm tied to 3.5). But I still wonder if there's a pretty straight forward implementation of calculating the average of integers without an external library. Do you happen to know about such implementation? Thanks!!

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• n & x commands&creating pointer&with using malloc [closed]

  - by gcc

  input 23 3 4 4 42 n 23 0 9 9 n n n 3 9 9 x //according to input,i should create int pointer arrays. pointer arrays // starting from 1 (that is initial arrays is arrays[1].when program sees n ,it // must be jumb to arrays 2 // the first int input 23 is num_arrays which used in malloc(sizeof(int)*num_arrays expected output arrays[1] 3 4 5 42 arrays[2] 23 0 9 9 arrays[5] 3 9 9 another input 12 2 3 4 n n 2 3 4 n 12 3 x expected output arrays[1] 2 3 4 arrays[3] 2 3 4 arrays[4] 12 3 x is stopper n is comman to create new pointer array i am new in this site anyone help me how can i write

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• C# rounding DateTime objects

  - by grenade

  I want to round dates/times to the nearest interval for a charting application. I'd like an extension method signature like follows so that the rounding can be acheived for any level of accuracy: static DateTime Round(this DateTime date, TimeSpan span); The idea is that if I pass in a timespan of ten minutes, it will round to the nearest ten minute interval. I can't get my head around the implementation and am hoping one of you will have written or used something similar before. I think either a floor, ceiling or nearest implementation is fine. Any ideas? Edit: Thanks to @tvanfosson & @ShuggyCoUk, the implementation looks like this: public static class DateExtensions { public static DateTime Round(this DateTime date, TimeSpan span) { long ticks = (date.Ticks / span.Ticks) + (span.Ticks / 2) + 1; return new DateTime(ticks * span.Ticks); } public static DateTime Floor(this DateTime date, TimeSpan span) { long ticks = (date.Ticks / span.Ticks); return new DateTime(ticks * span.Ticks); } public static DateTime Ceil(this DateTime date, TimeSpan span) { long ticks = (date.Ticks + span.Ticks - 1) / span.Ticks; return new DateTime(ticks * span.Ticks); } } And is called like so: DateTime nearestHour = DateTime.Now.Round(new TimeSpan(1,0,0)); DateTime minuteCeiling = DateTime.Now.Ceil(new TimeSpan(0,1,0)); DateTime weekFloor = DateTime.Now.Floor(new TimeSpan(7,0,0,0)); ... Cheers!

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• reopen or read and say why not reopened [closed]

  - by gcc

  input 23 3 4 4 42 n 23 0 9 9 n n n 3 9 9 x //according to input,i should create int pointer arrays. pointer arrays // starting from 1 (that is initial arrays is arrays[1].when program sees n ,it // must be jumb to arrays 2 // the first int input 23 is num_arrays which used in malloc(sizeof(int)*num_arrays expected output: elements of arrays[1] 3 4 5 42 elements of arrays[2] 23 0 9 9 elements of arrays[5] 3 9 9 another input 12 2 3 4 n n 2 3 4 n 12 3 x expected output elements of arrays[1] 2 3 4 elements of arrays[3] 2 3 4 elements of arrays[4] 12 3 specification: x is stopper n is comman to create new pointer array i am new in this site anyone help me how can i write

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• Quick Sort Vs Merge Sort

  - by Ganesh M

  Why might quick sort be better than merge sort ?

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• Using logarithms to normalize a vector to avoid overflow

  - by muscicapa

  http://stackoverflow.com/questions/2293762/problem-with-arithmetic-using-logarithms-to-avoid-numerical-underflow-take-2 Having seen the above and having seen softmax normalization I was trying to normalize a vector while avoiding overflow - that is (x1 x2 x3 x4 ... xn) the normalized form for me has the sum of squares as 1.0 So what I thought of doing is s=(2*log(x1)+2*log(x2)+...+2*log(xn))/2 so the two factor can be taken off and finally the normalized vector is exp(log(x1)-s), , ..., exp(log(xn)-s) but I am evidently doing something wrong here, what?

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• How do I convert decimal numbers to binary in Perl?

  - by David

  I am trying to make a program that converts decimal numbers or text into binary numbers in perl. The program asks for user input of a character or string , and then prints out the result to the console. How do I do this? My code I have been working on is below, but i cannot seem to fix it. print "Enter a number to convert: "; chomp($decimal = <STDIN>); print "\nConverting $number to binary...\n"; $remainder = $decimal%2; while($decimal > 0) { $decimal/2; print $remainder; }

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• What problems have you solved using artificial neural networks?

  - by knorv

  I'd like to know about specific problems you - the SO reader - have solved using artificial neural network techniques and what libraries/frameworks you used if you didn't roll your own. Questions: What problems have you used artificial neural networks to solve? What libraries/frameworks did you use? I'm looking for first-hand experiences, so please do not answer unless you have that.

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• C# average function without overflow exception

  - by Ron Klein

  .NET Framework 3.5. I'm trying to calculate the average of some pretty large numbers. For instance: using System; using System.Linq; class Program { static void Main(string[] args) { var items = new long[] { long.MaxValue - 100, long.MaxValue - 200, long.MaxValue - 300 }; try { var avg = items.Average(); Console.WriteLine(avg); } catch (OverflowException ex) { Console.WriteLine("can't calculate that!"); } Console.ReadLine(); } } Obviously, the mathematical result is 9223372036854775607 (long.MaxValue - 200), but I get an exception there. This is because the implementation (on my machine) to the Average extension method, as inspected by .NET Reflector is: public static double Average(this IEnumerable<long> source) { if (source == null) { throw Error.ArgumentNull("source"); } long num = 0L; long num2 = 0L; foreach (long num3 in source) { num += num3; num2 += 1L; } if (num2 <= 0L) { throw Error.NoElements(); } return (((double) num) / ((double) num2)); } I know I can use a BigInt library (yes, I know that it is included in .NET Framework 4.0, but I'm tied to 3.5). But I still wonder if there's a pretty straight forward implementation of calculating the average of integers without an external library. Do you happen to know about such implementation? Thanks!! UPDATE: The previous example, of three large integers, was just an example to illustrate the overflow issue. The question is about calculating an average of any set of numbers which might sum to a large number that exceeds the type's max value. Sorry about this confusion. I also changed the question's title to avoid additional confusion. Thanks all!!

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• Which data structure used in MySql?

  - by SomeUser

  Which data structure used in MySql?

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