牛客 u's的影响力（费马小定理、矩阵快速幂）

题目链接：点击这里

fib数列和g数列可以用 $O(logn)$ 矩阵快速幂求解。

根据费马小定理，由于 1e9+7 是素数，有 ${a\mathop{{}}\nolimits^{{1e9+6{\text{ }}}} \equiv 1 \left( mod\text{ }1e9+7 \right) }$ 。因此幂的指数对1e9+6取模即可。

(a + b) % p = (a % p + b % p) % p （1）

(a - b) % p = (a % p - b % p) % p （2）

(a * b) % p = (a % p * b % p) % p （3）

a^b % p = ((a % p)^b) % p （4）

#include<iostream>
#include<algorithm>
#include<string>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<stack>
#include<queue>
#include<map>
#include<set>

using namespace std;
typedef long long ll;
const ll mod = 1e9+6;

struct matrix {
    ll m[3][3];
};

matrix Mul(matrix A, matrix B)
{
    matrix ret;
    for(int i = 0; i < 3; i++)			//枚举行
    {
    	for(int j = 0; j < 3; j++)		//枚举列
        {
            ret.m[i][j] = 0;
            for(int k = 0; k < 3; k++)
                ret.m[i][j] = (ret.m[i][j] + A.m[i][k] * B.m[k][j]) % mod;  //取模1e9+6
        }
	}
    return ret;
}

matrix Pow(matrix A, ll n)
{
    matrix ret;
    ret.m[0][0] = 1;		//单位矩阵 
    ret.m[0][1] = 0;
    ret.m[0][2] = 0;
    ret.m[1][0] = 0;
    ret.m[1][1] = 1;
    ret.m[1][2] = 0;
    ret.m[2][0] = 0;
    ret.m[2][1] = 0;
    ret.m[2][2] = 1;
    while(n)
    {
        if(n&1)
            ret = Mul(ret,A);
        A = Mul(A,A);
        n>>=1;
    }
    return ret;
}

ll g(ll n)								//求g数列的第n项
{
	matrix ans, A;
    A.m[0][0] = 1;
    A.m[0][1] = 1;
    A.m[0][2] = 1;
	A.m[1][0] = 1;
    A.m[1][1] = 0;
    A.m[1][2] = 0;
    A.m[2][0] = 0;
    A.m[2][1] = 0;
    A.m[2][2] = 1;
    
    ans = Pow(A, n-1);
    
    return (ans.m[0][0] + ans.m[0][2]) % mod;		//取模1e9+6 
}

matrix fibMul(matrix A, matrix B)
{
    matrix ret;
    for(int i = 0; i < 2; i++)			//枚举行
    {
    	for(int j = 0; j < 2; j++)		//枚举列
        {
            ret.m[i][j] = 0;
            for(int k = 0; k < 2; k++)
                ret.m[i][j] = (ret.m[i][j] + A.m[i][k] * B.m[k][j]) % mod;  //取模1e9+6
        }
	}
    return ret;
}

matrix fibPow(matrix A, ll n)
{
    matrix ret;
    ret.m[0][0] = 1;		//单位矩阵 
    ret.m[0][1] = 0;
    ret.m[1][0] = 0;
    ret.m[1][1] = 1;
    while(n)
    {
        if(n&1)
            ret = fibMul(ret,A);
        A = fibMul(A,A);
        n>>=1;
    }
    return ret;
}

ll fib(ll n)							//求斐波那契数列的第n项
{
	matrix ans, A;
    A.m[0][0] = 1;
    A.m[0][1] = 1;
    A.m[1][0] = 1;
    A.m[1][1] = 0;
    
    ans = fibPow(A, n-1);
    
    return ans.m[0][0] % mod;		//取模1e9+6
}

ll power(ll a, ll b, ll m)
{
    ll res = 1;
 
    while(b)
	{
        if(b&1)
			res = res*a%m;
        a = a*a%m;
        b >>= 1;
    }
    return res;
}

int main()
{
	ll m = 1e9+7;
    ll n, x, y, a, b;
    cin>>n>>x>>y>>a>>b;
    if(n==1)
	{
		cout<<x%m<<endl;
		return 0;
	}
    if(n==2)
	{
		cout<<y%m<<endl;
		return 0;
	}
	
	if(x%m==0||y%m==0||a%m==0)
	{
		cout<<0;
		return 0;
	}
	
	a = power(a%m, b, m);
	
	cout<<power(x%m, fib(n-2), m)%m * power(y%m, fib(n-1), m)%m * power(a%m, g(n-2), m)%m<<endl;
	return 0;
}