HDU 2685 GCD推导

求$(a^n-1,a^m-1) \mod k$，自己手推，或者直接引用结论$(a^n-1,a^m-1) \equiv a^{(n,m)}-1 \mod k$

/** @Date    : 2017-09-21 21:41:26
  * @FileName: HDU 2685 结论 定理 推导.cpp
  * @Platform: Windows
  * @Author  : Lweleth (SoungEarlf@gmail.com)
  * @Link    : https://github.com/
  * @Version : $Id$
  */
#include <bits/stdc++.h>
#define LL long long
#define PII pair<int ,int>
#define MP(x, y) make_pair((x),(y))
#define fi first
#define se second
#define PB(x) push_back((x))
#define MMG(x) memset((x), -1,sizeof(x))
#define MMF(x) memset((x),0,sizeof(x))
#define MMI(x) memset((x), INF, sizeof(x))
using namespace std;

const int INF = 0x3f3f3f3f;
const int N = 1e5+20;
const double eps = 1e-8;

LL fpow(LL a, LL n, LL mod)
{
	LL res = 1;
	while(n)
	{
		if(n & 1)
			res = res * a % mod;
		a = a * a % mod;
		n >>= 1;
	}
	return res;
}

int main()
{
	int T;
	cin >> T;
	while(T--)
	{
		LL a, m, n, k;
		scanf("%lld%lld%lld%lld", &a, &m, &n, &k);
		printf("%lld\n", (fpow(a,__gcd(m,n),k) - 1 + k) % k);
	}
    return 0;
}