Codeforces Round #426 (Div. 2) C. The Meaningless Game (二分 or pow函数)

Posted 2020-09-30 Wally的博客

 

Slastyona and her loyal dog Pushok are playing a meaningless game that is indeed very interesting.

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 was recorded. She managed to recall the final results for each games, though, but all of her memories of them are vague. Help Slastyona verify their correctness, or, to put it another way, for each given pair of scores determine whether it was possible for a game to finish with such result or not.

Input

In the first string, the number of games n (1?≤?n?≤?350000) is given.

Each game is represented by a pair of scores a, b (1?≤?a,?b?≤?10⁹) – the results of Slastyona and Pushok, correspondingly.

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

Output

Yes
Yes
Yes
No
No
Yes

Note

First game might have been consisted of one round, in which the number 2 would have been chosen and Pushok would have won.

The second game needs exactly two rounds to finish with such result: in the first one, Slastyona would have said the number 5, and in the second one, Pushok would have barked the number 3.

 

 

题目大意：给你两个初值为1的数，如果你让其中一个数乘以x，则另一个数要乘以x的平方，问这样的操作若干次能否得到a和b？

 

解题思路：刚开始就只是想到了使用搜索的方法进行暴力求解，但是看了一下数据发现这样是绝对会超时的就果断放弃了。没想法看了别人的博客发现，a*b一定是一个数的三次方并且a和b取余这个数都是为0的，只要找到这个数就能进行判断了。

 

有两种做法：

1.二分：

 1 #include <iostream>
 2 #include <bits/stdc++.h>
 3 #define LL long long
 4 using namespace std;
 5 
 6 int main()
 7 {
 8     LL a,b;
 9     int t;
10     while(~scanf("%d",&t))
11     {
12         while(t--)
13         {
14             scanf("%I64d%I64d",&a,&b);
15             LL l=1,r=1000005,mid;
16             int flag=0;
17             while(l<=r)
18             {
19                 mid=(l+r)/2;
20                 //printf("%lld\n",mid);
21                 if(mid*mid*mid==a*b&&a%mid==0&&b%mid==0)
22                 {
23                     printf("Yes\n");
24                     flag=1;
25                     break;
26                 }
27                 else if(mid*mid*mid<a*b)
28                 {
29                     l=mid+1;
30                 }
31                 else
32                 {
33                     r=mid-1;
34                 }
35             }
36             if(!flag)
37             {
38                 printf("No\n");
39             }
40         }
41     }
42     return 0;
43 }

View Code

 

2.pow()+round()函数

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <algorithm>
 5 #include <bits/stdc++.h>
 6 using namespace std;
 7 
 8 int main()
 9 {
10     double a,b;
11     int t;
12     while(~scanf("%d",&t))
13     {
14         while(t--)
15         {
16           scanf("%lf%lf",&a,&b);
17           long long x,y,z;
18           double c=round(pow(a*b,1.0/3));
19           x=a,y=b,z=c;
20           if(x%z==0&&y%z==0&&z*z*z==x*y)
21           printf("Yes\n");
22           else
23           printf("No\n");
24         }
25     }
26     return 0;
27 }

View Code

知识点：

ceil(x)返回不小于x的最小整数值（然后转换为double型）。
floor(x)返回不大于x的最大整数值。
round(x)返回x的四舍五入整数值。

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