hdu 1005 Number Sequence 矩陣乘法

難度：3

[f(n) , f(n-1)] = [f(n-1) , f(n-2)] * [A,1;B,0]

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
const int N = 2;
int n , MOD;
struct Matrix {
    int n , m;
    int a[N][N];
    Matrix() {}
    Matrix(int _n,int _m):n(_n),m(_m){};
    void intput() {
        for(int i=0;i<n;i++)
        for(int j=0;j<m;j++)
            scanf("%d",&a[i][j]);
    }
    Matrix operator + (const Matrix &b) {
        Matrix tmp = Matrix(n,m);
        for(int i=0;i<n;i++)
        for(int j=0;j<m;j++)
            tmp.a[i][j] = a[i][j] + b.a[i][j];
        return tmp;
    }
    Matrix operator - (const Matrix &b) {
        Matrix tmp = Matrix(n,m);
        for(int i=0;i<n;i++)
        for(int j=0;j<m;j++)
            tmp.a[i][j] = a[i][j] - b.a[i][j];
        return tmp;
    }
    Matrix operator * (const Matrix &b) {
        Matrix tmp = Matrix(n,b.m);
        for(int i=0;i<n;i++)
        for(int j=0;j<b.m;j++)
            tmp.a[i][j] = 0;
        for(int i=0;i<n;i++)
        for(int j=0;j<b.m;j++)
        for(int k=0;k<m;k++) {
            tmp.a[i][j] += a[i][k]*b.a[k][j];
            tmp.a[i][j] %= MOD;
        }
        return tmp;
    }
};

Matrix operator ^ (Matrix a , int p) {
    Matrix ret = Matrix(a.n,a.m);
    for(int i=0;i<a.n;i++)
    for(int j=0;j<a.m;j++)
        ret.a[i][j] = (i == j ? 1 : 0);
    while(p) {
        if(p%2) ret = ret * a;
        a = a * a;
        p /= 2;
    }
    return ret;
}

Matrix A , B;

int a , b;
int main() {
    MOD = 7;
    while(~scanf("%d%d%d" , &a,&b,&n) && a+b+n) {
        if(n <= 2) {
            puts("1"); continue;
        }
        A = Matrix(1 , 2);
        B = Matrix(2 , 2);
        A.a[0][0] = A.a[0][1] = 1;
        B.a[0][0] = a; B.a[0][1] = 1;
        B.a[1][0] = b; B.a[1][1] = 0;
        A = A * (B^(n-2));
        int ans = A.a[0][0] % 7;
        printf("%d\n" , ans);
    }
    return 0;
}