第六届福建省大学生程序设计竞赛 Problem J RunningMan【数学】

朱雀 2022-07-31 09:56 119阅读 0赞

**Problem J RunningMan**

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Accept: 123 Submit: 551

Time Limit: 1000 mSec Memory Limit : 32768 KB

## ![prodesc.gif][] 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.

## ![prodesc.gif][] 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).

## ![prodesc.gif][] 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.)

## ![prodesc.gif][] Sample Input ##

100 100

200 100

## ![prodesc.gif][] Sample Output ##

Yes

## ![prodesc.gif][] 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.

解题思路：

几组数据可以得出，只要其中的最少的两组的人数大于另外一队则可以输出Yes。

#include<stdio.h>
    #include<string.h>
    #include<algorithm>
    using namespace std;
    int main()
    {
    	int T;
    	scanf("%d",&T);
    	while(T--)
    	{
    		int a,b;
    		scanf("%d%d",&a,&b);
    		int LL;
    		int LLL;
    		LLL=a%3;
    		LL=a/3;
    		if(LLL==0)
    		{
    			if(2*(LL+1)>b)
    			{
    				printf("Yes\n");
    			}
    			else
    			{
    				printf("No\n");
    			}
    		}
    		else if(LLL==1)
    		{
    			if(2*(LL+1)>b)
    			{
    				printf("Yes\n");
    			}
    			else
    			{
    				printf("No\n");
    			}
    		}
    		else if(LLL==2)
    		{
    			if(2*(LL+1)+1>b)
    			{
    				printf("Yes\n");
    			}
    			else
    			{
    				printf("No\n");
    			}
    		}
    	}
    	return 0;
    }

[prodesc.gif]: http://acm.fzu.edu.cn/image/prodesc.gif