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：

 1 #include <stdio.h>

 2 #include <algorithm>

 3 #include <iostream>

 4 #include <string.h>

 5 #include <string>

 6 #include <math.h>

 7 #include <stdlib.h>

 8 #include <queue>

 9 #include <stack>

10 #include <set>

11 #include <map>

12 #include <list>

13 #include <iomanip>

14 #include <vector>

15 #pragma comment(linker, "/STACK:1024000000,1024000000")

16 #pragma warning(disable:4786)

17 

18 using namespace std;

19 

20 const int INF = 0x3f3f3f3f;

21 const int MAX = 1000 + 10;

22 const double eps = 1e-8;

23 const double PI = acos(-1.0);

24 

25 int a[MAX];

26 

27 int main()

28 {

29     int T;

30     while(~scanf("%d",&T))

31     {

32         while(T --)

33         {

34             memset(a , 0 , sizeof(a));

35             int n;

36             cin >> n;

37             int ans = 0 , temp;

38             while(n --)

39             {

40                 scanf("%d",&temp);

41                 if(temp > 6000)

42                     ans ++;

43             }

44             cout << ans << endl;

45         }

46     }

47     return 0;

48 }

View Code