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.
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.
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).
6 2 4 75 45 8 8 16 16 247 994 1000000000 1000000
Yes Yes Yes No No Yes
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：放到给定一个a和b,对于他们每次的操作只能是a/k,b/k^2或者a/k^2,b/k。（每次都必须是整除）问最后他们是否可以同时变成1。
思路：假设满足条件，则必定存在s,p满足 s^2*p = a,p^2*s = b。如果确定一个s,那么p = a/s^2。p^2*s = a^2/s^3 = b。这是单调函数,所以，就可以使用二分。
#include<iostream>
#include<cmath>
#include<cstdio>
#define MAX 1000000
#define ll long long
using namespace std;
ll a,b;
bool Judge(ll num)
{
return a*a/num/num/num >= b;
}
int main( )
{
int n;
scanf("%d",&n);
while(n--){
scanf("%d%d",&a,&b);
ll left = 1,right = MAX;
while((right - left) > 1){
ll mid = left + ((right - left)>>1);
if(Judge(mid))
left = mid;
else
right = mid;
}
int s = left,p = a/left/left;
if(s*s*p == a && p*p*s == b)
printf("Yes\n");
else
printf("No\n");
}
}
优化思路：a = s^2*p,b = p^2*s。a*b = s^3*p^3。(s*p)^3 = a*b，同时 a%(s*p) == 0 && b%(s*p) == 0。所以，直接求出a*b的立方根即可。
#include<cmath>
#include<cstdio>
#define ll long long
using namespace std;
ll a,b;
int main( )
{
int n;
scanf("%d",&n);
while(n--){
scanf("%lld%lld",&a,&b);
ll c = pow(a*b,1.0/3) + 0.5;
puts(c*c*c==a*b && a%c==0 && b%c==0 ? "Yes":"No");
}
}
只不过因为从小数化正数的过程中会出现各种问题。所以，如果直接写的话不会AC。