摘要: Function Run FunTime Limit: 1000MS Memory Limit: 10000K Total Submissions: 14940 Accepted: 7736 DescriptionWe all love recursion! Don't we? Consider a three-parameter recursive function w(a, b, c): if a 20 or b > 20 or c > 20, then w(a, b, c) returns: w(20, 20, 20) if a int s[22][22][22];i 阅读全文
posted @ 2013-08-12 20:52 龙腾四海365 阅读(131) 评论(0) 推荐(0) 编辑
摘要: Train Problem I Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)Total Submission(s) : 8 Accepted Submission(s) : 2Problem DescriptionAs the new term comes, the Ignatius Train Station is very busy nowadays. A lot of student want to get back to school by train(because the 阅读全文
posted @ 2013-08-12 20:26 龙腾四海365 阅读(105) 评论(0) 推荐(0) 编辑
摘要: Rails Time Limit : 2000/1000ms (Java/Other) Memory Limit : 20000/10000K (Java/Other)Total Submission(s) : 31 Accepted Submission(s) : 17Problem DescriptionThere is a famous railway station in PopPush City. Country there is incredibly hilly. The station was built in last century. Unfortunately, funds 阅读全文
posted @ 2013-08-12 20:25 龙腾四海365 阅读(194) 评论(0) 推荐(0) 编辑
摘要: A/B Time Limit : 1000/1000ms (Java/Other) Memory Limit : 32768/32768K (Java/Other)Total Submission(s) : 6 Accepted Submission(s) : 5Problem Description要求(A/B)%9973,但由于A很大,我们只给出n(n=A%9973)(我们给定的A必能被B整除,且gcd(B,9973) = 1)。Input数据的第一行是一个T,表示有T组数据。每组数据有两个数n(0 int main(){int N;scanf("%d",&N) 阅读全文
posted @ 2013-08-12 20:24 龙腾四海365 阅读(116) 评论(0) 推荐(0) 编辑
摘要: String Matching Time Limit : 2000/1000ms (Java/Other) Memory Limit : 20000/10000K (Java/Other)Total Submission(s) : 23 Accepted Submission(s) : 7Problem DescriptionIt's easy to tell if two words are identical - just check the letters. But how do you tell if two words are almost identical? And ho 阅读全文
posted @ 2013-08-12 20:23 龙腾四海365 阅读(173) 评论(0) 推荐(0) 编辑
摘要: 绝对值排序 Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)Total Submission(s) : 39 Accepted Submission(s) : 28Problem Description输入n(n#include #include #include int cmp(const void *a,const void *b){ return abs(*(int *)b) - abs(*(int *)a);}int main(){ int n; while(scanf(&quo 阅读全文
posted @ 2013-08-12 19:41 龙腾四海365 阅读(161) 评论(0) 推荐(0) 编辑