I - RunningMan

最新推荐文章于 2019-11-27 11:42:11 发布

原创最新推荐文章于 2019-11-27 11:42:11 发布 · 247 阅读

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本文探讨了一个基于RunningMan节目的游戏策略问题。两支队伍通过分配成员参与三轮比赛来决出胜负，每轮比赛中队员不可重复使用。文章提供了一种算法来判断RunningMan队伍是否能确保胜利，无论对手如何布置人员。

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ZB loves watching RunningMan! There's a game in RunningMan called 100 vs 100.

There are two teams, each of many people. There are 3 rounds of fighting, in each round the two teams send some people to fight. In each round, whichever team sends more people wins, and if the two teams send the same amount of people, RunningMan team wins. Each person can be sent out to only one round. The team wins 2 rounds win the whole game. Note, the arrangement of the fighter in three rounds must be decided before the whole game starts.

We know that there are N people on the RunningMan team, and that there are M people on the opposite team. Now zb wants to know whether there exists an arrangement of people for the RunningMan team so that they can always win, no matter how the opposite team arrange their people.

Input

The first line contains an integer T, meaning the number of the cases. 1 <= T <= 50.

For each test case, there's one line consists of two integers N and M. (1 <= N, M <= 10^9).

Output

For each test case, Output "Yes" if there exists an arrangement of people so that the RunningMan team can always win. "No" if there isn't such an arrangement. (Without the quotation marks.)

Sample Input

2
100 100
200 100

Sample Output

No
Yes

Hint

In the second example, the RunningMan team can arrange 60, 60, 80 people for the three rounds. No matter how the opposite team arrange their 100 people, they cannot win.

题目大意：两个队进行3场战斗，每场两个队都配一些人，每场比赛中，如果两个队人数相等，Running man赢，每个人只能参加一次，三局两赢制，不论Runningman怎么安排人，他们都能赢，则输出Yes，否则输出No。

#include<iostream>
#include<cstdio>
using namespace std;
int main()

{
    int T;
    scanf("%d", &T);
    while(T--)
    {
        int n,m,a,b,c,t;
        scanf("%d%d",&n,&m);
            a=n/3;
            b=a;
            t=n%3;
            if(t==2)
                b++;
            if(b+a+2>m)
            printf("Yes\n");
            else
            printf("No\n");

    }
    return 0;
}