題目連結：http://acm.hdu.edu.cn/showproblem.php?pid=3221
題解：首先很容易發現遞推公式fn=fn-1*fn-2;寫出前幾項a,b,a*b,a*b^2,a^2*b^3,a^3*b^5;易發現a,b的指數為斐波那契數列。但是當N大一點時，斐波那契數列便變得非常大。那麼此時得用擴充套件尤拉定理降冪。
a^b%p=a^(b%phi(p)+phi(p))%p   b>=phi(p)（不要求a與p互質）
特別要主要使用條件， b>=phi(p)
接著用矩陣快速冪求斐波那契數列，再快速冪求答案即可。

程式碼如下：

#include <bits/stdc++.h>
#define ll long long
using namespace std;
const int maxn = 1000000 + 10;
ll MOD = 1e9 + 7;
ll a,b,n;
ll phi[maxn];
struct Matrix{
	ll n, m;
	ll mat[3][3];
	
	Matrix(ll a = 0, ll b = 0):n(a),m(b) {memset(mat,0,sizeof(mat));}
	void init(ll a, ll b) {n = a;m = b;}
	Matrix operator *(const Matrix & x){
		Matrix ans;ans.init(n,x.m);
		for(int i = 1;i <= n;i++){
			for(int j = 1;j <= x.m;j++){
				for(int k = 1;k <= m;k++){
					ans.mat[i][j] += mat[i][k] * x.mat[k][j];
					if(ans.mat[i][j] > phi[MOD]) ans.mat[i][j] = ans.mat[i][j] % phi[MOD] + phi[MOD];
				}
			}
		}
		return ans;
	}
};
Matrix q_pow(Matrix x, ll k){
	Matrix ret;ret.init(2,2);
	ret.mat[1][1] = ret.mat[2][2] = 1;
	while(k > 0){
		if(k & 1) ret = ret * x;
		x = x * x;
		k >>= 1;
	}
	return ret;
}
ll q_pow(ll x, ll k){
	ll ret = 1;
	while(k > 0){
		if(k & 1) ret = ret * x % MOD;
		x = x * x % MOD;
		k >>= 1;
	}
	return ret;
}
int primes[maxn],pcnt;
void get_phi(int n){
	memset(phi,0,sizeof(phi));
	phi[1] = 1;int t = 0;
	for(int i = 2;i <= n;i++){
		if(!phi[i]) {
			primes[++pcnt] = i;
			phi[i] = i - 1;
		}
		for(int j = 1;j <= pcnt;j++){
			t = primes[j];
			if(i * t > n) break;
			if(i % t == 0){
				phi[i * t] = phi[i] * t;
				break;
			}
			else phi[i * t] = phi[i] * (t - 1);
		}
	}
}
int main()
{
	get_phi(maxn - 10);
	int kase;scanf("%d",&kase);
	int T = 0;
	while(kase--){
		printf("Case #%d: ",++T);
		scanf("%I64d%I64d%I64d%I64d",&a,&b,&MOD,&n);
		if(MOD == 1) {printf("0\n");continue;}
		if(n == 1) {printf("%I64d\n",a % MOD);continue;}
		if(n == 2) {printf("%I64d\n",b % MOD);continue;}
		Matrix tmp;tmp.init(2,2);
		tmp.mat[1][1] = 0;tmp.mat[1][2] = 1;
		tmp.mat[2][1] = 1;tmp.mat[2][2] = 1;
		Matrix p = Matrix(1,2);	
		p.mat[1][1] = 1;p.mat[1][2] = 1;
		tmp = q_pow(tmp, n - 3);
		p = p * tmp;		ll ans = 1;
		ans *= q_pow(a,p.mat[1][1]);
		ans *= q_pow(b,p.mat[1][2]);
		ans %= MOD;
		printf("%I64d\n",ans);
	}
	return 0;
}