第十二届浙江省大学生程序设计大赛-Demacia of the Ancients
2015-06-26 14:39
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Demacia of the Ancients
Time Limit: 2 Seconds Memory Limit: 65536 KB
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
Time Limit: 2 Seconds Memory Limit: 65536 KB
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
#include <bits/stdc++.h> using namespace std; int main() { int t,n,i,sum; int a[100]; scanf("%d",&t); while(t--) { scanf("%d",&n); sum=0; for(i=0; i<n; i++) { scanf("%d",&a[i]); if(a[i]>6000) sum++; } cout<<sum<<endl; } return 0; }
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