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• expand a varchar column very slowly , why?

  - by francs

  Hi We need to modify a column of a big product table , usually normall ddl statments will be excutely fast ,but the above ddl statmens takes about 10 minnutes?I wonder know the reason! I just want to expand a varchar column?The following is the detailsl --table size wapreader_log= select pg_size_pretty(pg_relation_size('log_foot_mark')); pg_size_pretty ---------------- 5441 MB (1 row) --table ddl wapreader_log= \d log_foot_mark Table "wapreader_log.log_foot_mark" Column | Type | Modifiers -------------+-----------------------------+----------- id | integer | not null create_time | timestamp without time zone | sky_id | integer | url | character varying(1000) | refer_url | character varying(1000) | source | character varying(64) | users | character varying(64) | userm | character varying(64) | usert | character varying(64) | ip | character varying(32) | module | character varying(64) | resource_id | character varying(100) | user_agent | character varying(128) | Indexes: "pk_log_footmark" PRIMARY KEY, btree (id) --alter column wapreader_log= \timing Timing is on. wapreader_log= ALTER TABLE wapreader_log.log_foot_mark ALTER column user_agent TYPE character varying(256); ALTER TABLE Time: 603504.835 ms

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• ICU MessageFormat Currency Value Precison Lost

  - by Travis

  This may be a niche question but I'm working with ICU to format currency strings. I've bumped into a situation that I don't quite understand. When using the MesssageFormat class, why for a certain locale (Korea "ko"KR" for example) does it round currency values (e.g. 100.50 becomes ?101). For most locales (such as the US "en_US"), the precision of the argument passed in remains untouched (e.g. 100.50 becomes $100.50). I thought this might be a default rounding issue that some locales have (Swiss Francs "fr_CH" for example have a default 0.05 rounding) but South Korea "ko_KR" has none. Any ideas?

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• How to find minimum weight with maximum cost in 0-1 Knapsack algorithm?

  - by Nitin9791

  I am trying to solve a spoj problem Party Schedule the problem statement is- You just received another bill which you cannot pay because you lack the money. Unfortunately, this is not the first time to happen, and now you decide to investigate the cause of your constant monetary shortness. The reason is quite obvious: the lion's share of your money routinely disappears at the entrance of party localities. You make up your mind to solve the problem where it arises, namely at the parties themselves. You introduce a limit for your party budget and try to have the most possible fun with regard to this limit. You inquire beforehand about the entrance fee to each party and estimate how much fun you might have there. The list is readily compiled, but how do you actually pick the parties that give you the most fun and do not exceed your budget? Write a program which finds this optimal set of parties that offer the most fun. Keep in mind that your budget need not necessarily be reached exactly. Achieve the highest possible fun level, and do not spend more money than is absolutely necessary. Input The first line of the input specifies your party budget and the number n of parties. The following n lines contain two numbers each. The first number indicates the entrance fee of each party. Parties cost between 5 and 25 francs. The second number indicates the amount of fun of each party, given as an integer number ranging from 0 to 10. The budget will not exceed 500 and there will be at most 100 parties. All numbers are separated by a single space. There are many test cases. Input ends with 0 0. Output For each test case your program must output the sum of the entrance fees and the sum of all fun values of an optimal solution. Both numbers must be separated by a single space. Example Sample input: 50 10 12 3 15 8 16 9 16 6 10 2 21 9 18 4 12 4 17 8 18 9 50 10 13 8 19 10 16 8 12 9 10 2 12 8 13 5 15 5 11 7 16 2 0 0 Sample output: 49 26 48 32 now I know that it is an advance version of 0/1 knapsack problem where along with maximum cost we also have to find minimum weight that is less than a a given weight and have maximum cost. so I have used dp to solve this problem but still get a wrong awnser on submission while it is perfectly fine with given test cases. My code is typedef vector<int> vi; #define pb push_back #define FOR(i,n) for(int i=0;i<n;i++) int main() { //freopen("input.txt","r",stdin); while(1) { int W,n; cin>>W>>n; if(W==0 && n==0) break; int K[n+1][W+1]; vi val,wt; FOR(i,n) { int x,y; cin>>x>>y; wt.pb(x); val.pb(y); } FOR(i,n+1) { FOR(w,W+1) { if(i==0 || w==0) { K[i][w]=0; } else if (wt[i-1] <= w) { if(val[i-1] + K[i-1][w-wt[i-1]]>=K[i-1][w]) { K[i][w]=val[i-1] + K[i-1][w-wt[i-1]]; } else { K[i][w]=K[i-1][w]; } } else { K[i][w] = K[i-1][w]; } } } int a1=K[n][W],a2; for(int j=0;j<W;j++) { if(K[n][j]==a1) { a2=j; break; } } cout<<a2<<" "<<a1<<"\n"; } return 0; } Could anyone suggest what am I missing??

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