D - 下个也是签到题 FZU - 2221(博弈)

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.

 

 

AC代码：

 1 #include<stdio.h>  
 2 #include<string.h>  
 3 #include<algorithm>  
 4 using namespace std;  
 5 int main()  
 6 {  
 7     int t;  
 8     int n,m;  
 9     scanf("%d",&t);  
10     while(t--)  
11     {  
12         scanf("%d%d",&n,&m);  
13         if((n+1)*1.0>=(m*3/2))  
14             printf("Yes\n");  
15         else  
16             printf("No\n");  
17     }  
18     return 0;  
19 }

 

 

和队友在那看了一个多小时，各种提交各种错，看过别人的代码后发现是少乘了个1.0真是无语了。

 

转载于:https://www.cnblogs.com/ruruozhenhao/p/8908180.html