HDU 4549 M斐波那契数列（矩阵快速幂）

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
#include<string>
#include<iostream>
#include<queue>
#include<cmath>
#include<map>
#include<stack>
#include<bitset>
using namespace std;
#define REPF( i , a , b ) for ( int i = a ; i <= b ; ++ i )
#define REP( i , n ) for ( int i = 0 ; i < n ; ++ i )
#define CLEAR( a , x ) memset ( a , x , sizeof a )
typedef long long LL;
typedef pair<int,int>pil;
const int INF = 0x3f3f3f3f;
const int MOD=1e9+6;
LL a,b,n;
struct Matrix{
    LL mat[2][2];
    void Clear()
    {
        CLEAR(mat,0);
    }
};
Matrix mult(Matrix m1,Matrix m2)
{
    Matrix ans;
    for(int i=0;i<2;i++)
        for(int j=0;j<2;j++)
        {
            ans.mat[i][j]=0;
            for(int k=0;k<2;k++)
                ans.mat[i][j]=(ans.mat[i][j]+m1.mat[i][k]*m2.mat[k][j])%MOD;
        }
    return ans;
}
Matrix Pow(Matrix m1,LL b)
{
    Matrix ans;ans.Clear();
    for(int i=0;i<2;i++)
        ans.mat[i][i]=1;
    while(b)
    {
        if(b&1)
            ans=mult(ans,m1);
        b>>=1;
        m1=mult(m1,m1);
    }
    return ans;
}
LL quick_mod(LL a,LL b)
{
    LL ans=1;
    while(b)
    {
        if(b&1)
            ans=ans*a%1000000007;
        b>>=1;
        a=a*a%1000000007;
    }
    return ans;
}
int main()
{
    while(~scanf("%lld%lld%lld",&a,&b,&n))
    {
         Matrix A;
         if(n<=1)
         {
             printf("%lld\n",n==0?a:b);
             continue;
         }
         A.mat[0][0]=A.mat[0][1]=1;
         A.mat[1][0]=1;A.mat[1][1]=0;
         A=Pow(A,n-1);
         LL m,k;
         m=(A.mat[0][0])%MOD;k=(A.mat[0][1])%MOD;
         LL ans=1;
         ans=ans*quick_mod(a,k)%1000000007;
         ans=ans*quick_mod(b,m)%1000000007;
         printf("%lld\n",ans);
    }
    return 0;
}