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 = 1000 + 10;
const double eps = 1e-8;
const double PI = acos(-1.0);

int a[MAX];

int main()
{
    int T;
    while(~scanf("%d",&T))
    {
        while(T --)
        {
            memset(a , 0 , sizeof(a));
            int n;
            cin >> n;
            int ans = 0 , temp;
            while(n --)
            {
                scanf("%d",&temp);
                if(temp > 6000)
                    ans ++;
            }
            cout << ans << endl;
        }
    }
    return 0;
}