HDU5873

Football Games

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 514    Accepted Submission(s): 188

Problem Description

A mysterious country will hold a football world championships---Abnormal Cup, attracting football teams and fans from all around the world. This country is so mysterious that none of the information of the games will be open to the public till the end of all the matches. And finally only the score of each team will be announced. 
  
  At the first phase of the championships, teams are divided into M groups using the single round robin rule where one and only one game will be played between each pair of teams within each group. The winner of a game scores 2 points, the loser scores 0, when the game is tied both score 1 point. The schedule of these games are unknown, only the scores of each team in each group are available.
  
  When those games finished, some insider revealed that there were some false scores in some groups. This has aroused great concern among the pubic, so the the Association of Credit Management (ACM) asks you to judge which groups' scores must be false.

 

 

Input

Multiple test cases, process till end of the input.
  
  For each case, the first line contains a positive integers M, which is the number of groups.
  The i-th of the next M lines begins with a positive integer Bi representing the number of teams in the i-th group, followed by Bi nonnegative integers representing the score of each team in this group.


number of test cases <= 10
M<= 100
B[i]<= 20000
score of each team <= 20000

 

 

Output

For each test case, output M lines. Output ``F" (without quotes) if the scores in the i-th group must be false, output ``T" (without quotes) otherwise. See samples for detail.

 

 

Sample Input

2

3 0 5 1

2 1 1

 

Sample Output

F

T

 

签到水题，只需要判断两个条件：

1.分数之和==2×C(2,n)，因为只要比一场赛，无论结果，双方总得分和为2。

2.分数最高的不会超过2×(n-1)分，即假设他与其他n-1支队伍踢都赢，最多只能有2×(n-1)分

//2016.9.11
#include <iostream>
#include <cstdio>
#define N 20005

using namespace std;

int b[N];

int C(int m, int n)
{
    if(n-m < m)m = n-m;
    int ans = 1;
    for(int i = 0; i < m; i++)
    {
        ans *= n;
        n--;
    }
    for(int i = 2; i <= m; i++)
      ans /= i;
    return ans;
}

int main()
{
    int m, n;
    while(scanf("%d", &m)!=EOF)
    {
        while(m--)
        {
            scanf("%d", &n);
            int sum = 0, max = 0;
            for(int i = 0; i < n; i++)
            {
                scanf("%d", &b[i]);
                sum += b[i];
                if(max<b[i])max = b[i];
            }
            int c = C(2, n);
            if(sum==2*c && max<=2*(n-1))
                  printf("T\n");
            else printf("F\n");
        }
    }

    return 0;
}