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.

, 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

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.

题意：

m个小组，每个小组n支队伍进行比赛，任意两支队伍之间有一场比赛

一场比赛里赢得+2分输的+0分，打平的话每队+1分

先给出每支队伍的得分，判断这些得分是否满足小组比赛的条件

思路：根据这个比赛规则，我们可以发现，每场比赛都有2个积分会出去。

那么问题就很好解决了，先要给那些

得分从小到大排序，对于当前i，与之前的i-1支队伍比赛完之后，所有的比赛的总得分至少是(i-1)*i（因为这i只队伍还要和第i+1~n的队伍打比赛，也可能获得分数

#include<bits/stdc++.h>

using namespace std;

const int maxn = +;

int n,a[maxn];

int main()

{

    int T;

    while(scanf("%d",&T)!=EOF)

    {

        while(T--)

        {

            int res = ;

            scanf("%d",&n);

            for(int i = ;i<=n;i++)

                scanf("%d",&a[i]),res+=a[i];

            sort(a+,a++n);

            int sum = ;

            int flag = ;

            for(int i = ;i<=n;i++)

            {

                sum+=a[i];

                if(sum<i*(i-))

                {

                    flag=;

                    break;

                }

            }

            if(res!=n*(n-))

                flag=;

            if(flag)

                printf("F\n");

            else

                printf("T\n");

        }

    }

}  

）；

最后只要算出比赛的场次*2==总分就可以。

一共有n组，那么任意两只队伍要比赛的话排列组合共有Cn^2种情况，即n*(n-1)/2;