M斐波那契数列

题目描述：
M斐波那契数列F[n]是一种整数数列，它的定义如下：
F[0] = a
F[1] = b
F[n] = F[n-1] * F[n-2] ( n > 1 )
现在给出a, b, n，你能求出F[n]的值吗？（ 0 <= a, b, n <= 10^9 ）
由于F[n]可能很大，你只需输出F[n]对1000000007取模后的值即可

题解：
列出F[n]的前几项会发现，F[n] = a^f(n-2) * b^f(n-1)，f(n)为斐波拉契数列，用矩阵快速幂求解
费马小定理：a^k % p = a^(k % (p-1)) 其中，p 为质数
a^f(n-2) * b^f(n-1) % p = a^( f(n-2) % (p-1) ) * b^( f(n-1) % (p-1) ) % p

代码

#include <stdio.h>#include <stdlib.h>#include <string.h>#include <math.h>#include <time.h>#include <iostream>#include <algorithm>#include <string>#include <set>#include <map>#include <queue>#include <stack>#include <vector>using namespace std;const int mod = 1e9+7;int n,x,y;struct Matrix{    int n,m,d[5][5];    Matrix (int a,int b)    {        n = a, m = b;        memset(d,0,sizeof(d));    }    void Copy(int *tmp)    {        for (int i=0;i<n;i++)            for (int j=0;j<m;j++)            {                d[i][j] = (*tmp);                tmp++;            }    }    friend Matrix operator * (const Matrix &a,const Matrix &b)    {        Matrix c(a.n,b.m);        for (int i=0;i<a.n;i++)            for (int j=0;j<b.m;j++)            {                long long tmp = 0;                for (int k=0;k<a.m;k++)                {                    tmp += (long long)a.d[i][k] * b.d[k][j] % (mod-1);                    tmp %= (mod-1);                }                c.d[i][j] = tmp % (mod-1);            }        return c;    }};long long qpow(int x,int k){    long long ans = 1;    x %= mod;    while (k)    {        if (k & 1) ans = (ans*x) % mod;        x = (long long)x*x % mod;        k >>= 1;    }    return ans;}long long solve(int x,int y,int n){    if (n == -1) return x;    if (n == 0) return y;    Matrix A(1,2),B(2,2);    int a[2] = {1,0};    int b[4] = {1,1,1,0};    A.Copy(a);    B.Copy(b);    while (n)    {        if (n & 1) A = A*B;        B = B*B;        n >>= 1;    }    long long f=0,g=0;    f=qpow(x,A.d[0][1]);    g=qpow(y,A.d[0][0]);    return (f * g % mod);}int main(){    while (scanf("%d %d %d",&x,&y,&n)!=EOF)    {        printf("%I64d\n",solve(x,y,n-1));    }    return 0;}