codeforces 834 C 思路题

http://codeforces.com/problemset/problem/834/C

C. The Meaningless Game

time limit per test
1 second

memory limit per test
256 megabytes

input
standard input

output
standard output



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.

题意：现在两个人做游戏，每个人刚开始都是数字1，谁赢了就能乘以k^2，输的乘以k，现在给你最终这两个人的得分，让你判断是否有这个可能，有可能的话Yes，否则No。

思路：不论谁乘以k^2，谁乘以k，他们一定是都至少乘了一个k，并且他两个人的乘积一定是乘了一个k的三次方。

所以先把所有数字的三次方存一下，然后先判断这两个数字乘积是否为三次方的数字（x^3），若是的话，在判断这两个数字是否是这个数字的x的倍数。是的话为Yes，否则No。

代码：

#include<cstdio>
#include<string>
#include<iostream>
#include<map>
#include <algorithm>
#define LL long long
using namespace std;
map<LL,int>ma;
void init()
{
for(LL i=1; i<=1000000; i++)
ma[i*i*i]=i;
}
int main()
{
init();
int t;
LL a,b;
scanf("%d",&t);
while(t--)
{
scanf("%lld%lld",&a,&b);
int x=ma[a*b];
if(x&&a%x==0&&b%x==0)
printf("Yes\n");
else
printf("No\n");
}
}