C(clang 3.9) 解法, 执行用时: 1076ms, 内存消耗: 9596K, 提交时间: 2020-07-20 12:29:08

#include<stdio.h>
#include<math.h>
#define M (int)(1e9+7)
int gcd(int n);
int f(int n,int c)
{
	if(n==1) return 1;
	return 1ll*c*f(gcd(n),c)%M;
}
int gcd(int n)
{
	int i,head;
	head=sqrt(n);
	for(i=2;i<=head;i++) if(n%i==0) return n/i;
	return 1;
}

int main()
{
	int t,n,c,i;
	scanf("%d",&t);
	for(i=0;i<t;i++)
	{
		scanf("%d %d",&n,&c);
		printf("%d\n",f(n,c));
	}
}

C++14(g++5.4) 解法, 执行用时: 1165ms, 内存消耗: 9700K, 提交时间: 2020-07-25 20:09:30

#include<bits/stdc++.h>
typedef long long ll;
using namespace std;
const int M=1e9+7;
int main(){

	int t;
	scanf("%d",&t);
	while(t--){
		int n,c;
		ll ans=1;
		scanf("%d %d",&n,&c);
		for(int i=2;i*i<=n;i++){
			while(n%i==0){
				n/=i;
				ans=ans*c%M;
			}
		}
		if(n!=1) ans=ans*c%M;
		printf("%lld\n",ans);
	}
	return 0;
}

C++(clang++ 11.0.1) 解法, 执行用时: 1336ms, 内存消耗: 9696K, 提交时间: 2023-05-02 17:48:26

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll mod = 1e9+7;

int T;
int n,c;


int main(){
	scanf("%d",&T);
	while(T--){
		scanf("%d%d",&n,&c);
		ll ans=1;
		for(int i=2;i*i<=n;++i)
		while(n%i==0){
			n/=i;
			ans=ans*c%mod;
		}
		if(n>1)ans=ans*c%mod;
		printf("%lld\n",ans);
	}
}