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CodeForces - 834C

2022-08-10 12:24:00 【51CTO】


Slastyona and her loyal dog Pushok are playing a meaningless game

The game consists of multiple rounds. Its rules are very simple: in each round, a natural number k is chosen. Then, the one who says (or barks) it faster than the other wins the round. After that, the winner's score is multiplied by k2, and the loser's score is multiplied by k. In the beginning of the game, both Slastyona and Pushok have scores equal to one.

Unfortunately, Slastyona had lost her notepad where the history of all n games

Input

In the first string, the number of games n (1 ≤ n ≤ 350000)

Each game is represented by a pair of scores a, b (1 ≤ a, b ≤ 109)

Output

For each pair of scores, answer "Yes" if it's possible for a game to finish with given score, and "No" otherwise.

You can output each letter in arbitrary case (upper or lower).

Example

Input 



      
      

       
       6
       
       
2 4
       
       
75 45
       
       
8 8
       
       
16 16
       
       
247 994
       
       
1000000000 1000000
      
      

      
      
    
       
       
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    • 
       
       
  • 2.
    • 
       
       
  • 3.
    • 
       
       
  • 4.
    • 
       
       
  • 5.
    • 
       
       
  • 6.
    • 
       
       
  • 7.
    •



Output 

      
      

       
       Yes
       
       
Yes
       
       
Yes
       
       
No
       
       
No
       
       
Yes
      
      

      
      
    
       
       
  • 1.
    • 
       
       
  • 2.
    • 
       
       
  • 3.
    • 
       
       
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  • 6.
    •



题目大概：

两个人说数字，谁的大谁赢，赢的一方的分数乘大的数字的平方，输的分数乘大的数字，先知道两方分数，问比赛评分是否对。（比赛可以有多轮）

思路：

当一轮比赛，两方分数相乘为大的数的三次方，当多轮时，也成立，为多轮大数乘积的三次方。用二分求出最后的值，检验该值是否对，并输出结果。

代码：


      
      

       
       #include <iostream>
       
       

       
       #include <string>
       
       

       
       #include <queue>
       
       

       
       #include <cstring>
       
       

       
       #include <cstdio>
       
       

       
       using 
       
       namespace 
       
       std;
       
       

       
       
 
       
       long 
       
       long 
       
       go(
       
       long 
       
       long 
       
       sum)
       
       
{
       
       
   
       
       long 
       
       long 
       
       l,
       
       r;
       
       
   
       
       l
       
       =
       
       0;
       
       r
       
       =
       
       1000000;
       
       

       
       
   
       
       while(
       
       l
       
       !=
       
       r)
       
       
   {
       
       
       
       
       long 
       
       long 
       
       mid
       
       =(
       
       l
       
       +
       
       r
       
       +
       
       1)
       
       /
       
       2;
       
       

       
       
       
       
       if(
       
       mid
       
       *
       
       mid
       
       *
       
       mid
       
       >
       
       sum){
       
       
       
       
       r
       
       =
       
       mid
       
       -
       
       1;
       
       
       }
       
       
       
       
       else {
       
       
       
       
       l
       
       =
       
       mid;
       
       
       }
       
       
   }
       
       
   
       
       return 
       
       l;
       
       
}
       
       

       
       

       
       int 
       
       main()
       
       
{
       
       

       
       int 
       
       t;
       
       

       
       scanf(
       
       "%d",
       
       &
       
       t);
       
       

       
       for(
       
       int 
       
       i
       
       =
       
       0;
       
       i
       
       <
       
       t;
       
       i
       
       ++)
       
       
{
       
       
    
       
       long 
       
       long 
       
       n,
       
       m;
       
       
    
       
       scanf(
       
       "%I64d%I64d",
       
       &
       
       n,
       
       &
       
       m);
       
       

       
       

       
       
  
       
       long 
       
       long 
       
       g
       
       =
       
       go(
       
       n
       
       *
       
       m);
       
       

       
       

       
       
 
       
       if(
       
       g
       
       *
       
       g
       
       *
       
       g
       
       !=
       
       n
       
       *
       
       m)
       
       printf(
       
       "No\n");
       
       
    
       
       else 
       
       if(
       
       n
       
       %
       
       g
       
       ==
       
       0
       
       &&
       
       m
       
       %
       
       g
       
       ==
       
       0)
       
       printf(
       
       "Yes\n");
       
       
    
       
       else 
       
       printf(
       
       "No\n");
       
       

       
       

       
       

       
       
}
       
       
 
       
       return 
       
       0;
       
       

       
       

       
       

       
       
}
      
      

      
      
    
       
       
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  • 51.
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原网站

版权声明
本文为[51CTO]所创，转载请带上原文链接，感谢
https://blog.51cto.com/u_15747246/5563298

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