zoj The 12th Zhejiang Provincial Collegiate Programming Contest Demacia of the Ancients

http://acm.zju.edu.cn/onlinejudge/showContestProblem.do?problemId=5504

 The 12th Zhejiang Provincial Collegiate Programming Contest - L

Demacia of the Ancients

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Time Limit: 2 Seconds      Memory Limit: 65536 KB

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There is a popular multiplayer online battle arena game called Demacia of the Ancients. There are lots of professional teams playing this game. A team will be approved as Level K if there are exact K team members whose match making ranking (MMR) is strictly greater than 6000.

You are given a list of teams. Please calculate the level of each team.

Input

There are multiple test cases. The first line of input contains an integer T indicating the number of test cases. For each test case:

The first line contains an integer N (1 <= N <= 10) indicating the number of team members.

The second line contains N integers representing the MMR of each team member. All MMRs are non-negative integers less than or equal to 9999.

Output

For each test case, output the level of the given team.

Sample Input

3
5
7986 6984 6645 6200 6150
5
7401 7377 6900 6000 4300
3
800 600 200

Sample Output

5
3
0

分析；

这个也是签到题。
就是找超出6000的个数。

AC代码：

 #include <stdio.h>
 #include <algorithm>
 #include <iostream>
 #include <string.h>
 #include <string>
 #include <math.h>
 #include <stdlib.h>
 #include <queue>
 #include <stack>
 #include <set>
 #include <map>
 #include <list>
 #include <iomanip>
 #include <vector>
 #pragma comment(linker, "/STACK:1024000000,1024000000")
 #pragma warning(disable:4786)

 using namespace std;

 const int INF = 0x3f3f3f3f;
 const int MAX =  + ;
 const double eps = 1e-;
 const double PI = acos(-1.0);

 int a[MAX];

 int main()
 {
     int T;
     while(~scanf("%d",&T))
     {
         while(T --)
         {
             memset(a ,  , sizeof(a));
             int n;
             cin >> n;
             int ans =  , temp;
             while(n --)
             {
                 scanf("%d",&temp);
                 if(temp > )
                     ans ++;
             }
             cout << ans << endl;
         }
     }
     return ;
 }