poj 2685 GCD快速幂

自闭，因为gcd里面必须用快速幂优化，想不到有这么个推论：

gcd(am-1,an-1) = agcd(m,n)-1

推广:

若 gcd(a,b)=1

gcd(am-bm,an-bn) = agcd(m,n)-bgcd(m,n)

 

#include
#include
#include
#include
#define ll long long
using namespace std;
ll a,n,m,k;
ll power_mod(ll a,ll b,ll n)
{
	if (b==0) return 1;
	if (b==1) return a%n;
	ll tmp=power_mod(a,b/2,n);
	tmp=(tmp*tmp)%n;
	if (b&1)tmp=tmp*a%n;
	return tmp;
}
ll gcd(ll a,ll b)
{
	return b?gcd(b,a%b):a;
}
int main()
{
	int T;
	scanf("%d",&T);
	while (T--)
	{
		scanf("%lld%lld%lld%lld",&a,&m,&n,&k);
		ll t=gcd(n,m);
		ll ans=(power_mod(a,t,k)-1+k)%k;
		printf("%lld\n",ans);
	}
	return 0;
}