矩阵快速幂递推+欧拉降幂

#include <bits/stdc++.h>
#define mod 1000000006
#define mo 1000000007
#define PI acos(-1)
using namespace std;
typedef long long ll ;
struct node{
    ll a[3][3];
    node(){memset(a ,0 , sizeof(a));}
};

node mul(node A, node B)
{
    node C ;
    for(int i = 0 ; i < 3 ; i++)
        for(int j = 0 ; j < 3 ; j++)
            for(int k = 0 ; k < 3 ; k++)
                C.a[i][j] = (C.a[i][j] + A.a[i][k] * B.a[k][j]%mod)%mod;
    return C;
}

node quickpow(node A, ll b)
{
    node ans ;
    for(int i = 0 ; i < 3 ; i++) ans.a[i][i] = 1;
    while(b){if(b&1) ans=mul(ans,A);b>>=1,A=mul(A,A);}
    return ans;
}

ll quick(ll a , ll b)
{
    //if(a == 0) return 0 ;//避免0 的 0 次方为1。
    ll ans = 1 ;
    while(b){if(b&1) ans=ans*a % mo ; b>>=1 , a = a*a%mo;}
    return ans%mo;
}

int main()
{
    ll n , x , y , a , b ;
    cin >> n >> x >> y >> a >> b;
    x %= mo , y %= mo , a %= mo , b %= mod;
    if(n == 1) cout << x << endl;
    else if(n == 2) cout << y << endl;
    else if(n == 3)
    {
        if(x == 0 || y == 0 ||a == 0) cout << 0 << endl; // 注意特判 ，0的0次方快速幂为1
        cout << x * y % mo * quick(a , b) % mo;
    }
    else
    {
        if(x == 0 || y == 0 ||a == 0)//注意特判 ，0的0次方快速幂为1
        {
            cout << 0 << endl;
            return 0 ;
        }
        node A , B , C; ll m1, m2, m3 ;
        A.a[0][0] = 1 , A.a[0][1] = 1 , A.a[1][0] = 1 ;
        B.a[0][0] = 0 , B.a[1][0] = 1 ;
        C = mul(quickpow(A , n-2) , B);
        m1 = C.a[0][0] ;
        B.a[0][0] = 1 , B.a[1][0] = 0 ;
        C = mul(quickpow(A , n-2) , B);
        m2 = C.a[0][0] ;
        A.a[0][2] = 1 , A.a[2][2] = 1 ;
        B.a[2][0] = 1 ;
        C = mul(quickpow(A , n-3) , B);
        m3 = b * C.a[0][0] % mod ;
        cout << quick(x , m1) * quick(y , m2)%mo * quick(a , m3)%mo << endl;
    }

    return 0;
}