FZU Problem 2221 RunningMan

最新推荐文章于 2019-04-08 17:45:46 发布

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本文介绍了一个基于 Running Man 节目中 100vs100 游戏胜负判断的问题，并提供了一段 C 语言实现的代码来解决这个问题。游戏规则涉及两队在三轮比赛中派遣人员进行对抗，获胜条件为赢得两轮以上的队伍。

Accept: 840 Submit: 2574
Time Limit: 1000 mSec Memory Limit : 32768 KB

Problem Description

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

2100 100200 100

Sample Output

NoYes

虽然是到水题，但在当时也让我们想了很久....

AC代码：

#include<cstdio>
int main() 
{
	int t,n,m;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d %d",&n,&m);
		if((n+1)/3>=m/2)	puts("Yes");
		else				puts("No");
	}
	return 0;
}