Codeforces Round #426 (Div. 2)The Meaningless Game【数学题】【水题】-白红宇

Codeforces Round #426 (Div. 2)The Meaningless Game【数学题】【水题】

阅读量：6115 次

发布时间：2019-06-21

本文共 2651 字，大约阅读时间需要 8 分钟。

$\text{[math]}$

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 
62 475 458 816 16247 9941000000000 1000000
        output 
YesYesYesNoNoYes

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.

假设num1由a*b^2*c*d^2 则num2由a^2*b*c^2*d

则可求出c= 立方根（num1*num2）=a*b*c*d

则可由num1/num2*c 求出其中一组数的乘积的平方,num2/num1*c同理

从而num1=a*c*(b*d)^2,num2同理，只要验证能否组成num1和num2即可

ps(注意会爆int，用cin输入会超时)

#include 
    
     #include 
     
      #include 
      
       #include 
       
        #include 
        
         #include 
         
          #include 
          #include 
           
            #define INF 0x3f3f3f3f#define ms(x,y) memset(x,y,sizeof(x))using namespace std;typedef long long ll;const double pi = acos(-1);const double eps = 1e-8;const int maxn = 1e6 + 10;;int main(){ int n; scanf("%d", &n); for (int i = 0; i < n; i++) { ll a, b, c; scanf("%I64d%I64d", &a, &b); c = a*b; c = pow(c, 1.0 / 3) + 0.5; ll p, q; p = a*c / b; q = b*c / a; p = sqrt(p); q = sqrt(q); if (min(a, b) == min(p*q*q, p*p*q) && max(a, b) == max(p*q*q, p*p*q)) puts("YES"); else puts("NO"); } return 0;}

转载于:https://www.cnblogs.com/Archger/p/8451617.html

你可能感兴趣的文章