Shaolin Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)Total Submission(s): 4370 Accepted Submission(s): 1883 Problem Description Shaolin temple is very famous for its Kongfu monks.A lot of young men go to Shaol…
Shaolin Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)Total Submission(s): 315 Accepted Submission(s): 162 Problem Description Shaolin temple is very famous for its Kongfu monks.A lot of young men go to Shaolin…
水果 Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 15377 Accepted Submission(s): 5977 Problem Description 夏天来了~~好开心啊,呵呵,好多好多水果~~Joe经营着一个不大的水果店.他认为生存之道就是经营最受顾客欢迎的水果.现在他想要一份水果销售情况的明细表,这样Joe就可以很…
Shaolin Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others) Total Submission(s): 3191 Accepted Submission(s): 1350 Problem Description Shaolin temple is very famous for its Kongfu monks.A lot of young men go to Shao…
Shaolin Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)Total Submission(s): 64 Accepted Submission(s): 38 Problem Description Shaolin temple is very famous for its Kongfu monks.A lot of young men go to Shaolin t…
产生冠军 Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 26204 Accepted Submission(s): 11751 Problem Description 有一群人,打乒乓球比赛,两两捉对撕杀,每两个人之间最多打一场比赛.球赛的规则如下:如果A打败了B,B又打败了C,而A与C之间没有进行过比赛,那么就认定,A一定能打败…
项目管理 Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 730 Accepted Submission(s): 258 Problem Description 我们建造了一个大项目! 这个项目有n个节点,用非常多边连接起来.而且这个项目是连通的! 两个节点间可能有多条边,只是一条边的两端必定是不同的节点. 每一个节点都有一个能量…
Misha hacked the Codeforces site. Then he decided to let all the users change their handles. A user can now change his handle any number of times. But each new handle must not be equal to any handle that is already used or that was used at some point…
题目 题目 分析 自认已经很简洁了,虽说牺牲了一些效率 代码 #include <bits/stdc++.h> using namespace std; set <string> m; string s[120003]; int main() { int n; while(cin>>s[n]) m.insert(s[n++]); for(int i=0;i<n;i++) { int l=s[i].length(); for(int j=1;j<…
