小a与黄金街道（欧拉函数+快速幂）

链接：https://ac.nowcoder.com/acm/contest/317/D

本题的关键是所有质数的和 q  = n*euler(n)/2;

先用欧拉函数求出所有质数的和，再用快速幂求出k^q

#include
#include
#include
#include
using namespace std;

long long mod = 1e9+7;

long long f(long long k,long long p){
	long long res = 1;
	while(p){
		if(p&1) res = (res*k)%mod;
		k = (k*k)%mod;
		p >>= 1;
	}
	return res;
}

long long euler(long long n){
	long long res = n,a = n;
	for(long long i = 2;i*i<=a;++i){
		if(a % i == 0){
			res = res/i*(i-1);
			while(a % i == 0){
				a /= i;
			}
		}
	}
	if(a > 1) res = res/a*(a-1);
	return res;
}

int main(){
	long long n,k,a,b,p;
	cin>>n>>k>>a>>b;
	p = n*euler(n)/2;
	p %= mod-1;
	cout<<(a+b)*f(k,p)%mod<