Codeforces Round #257 (Div. 2) B. Jzzhu and Sequences (矩阵快速幂)

题目链接：http://codeforces.com/problemset/problem/450/B

题意很好懂，矩阵快速幂模版题。

/*
    | 1, -1 | | fn   |
    | 1,  0 | | fn-1 |
*/
#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
typedef __int64 LL;
LL mod = 1e9 + 7;
struct data {
    LL mat[3][3];
};

data operator* (data a , data b) {
    data res;
    for(int i = 1 ; i <= 2 ; i++) {
        for(int j = 1 ; j <= 2 ; j++) {
            res.mat[i][j] = 0;
            for(int k = 1 ; k <= 2 ; k++) {
                res.mat[i][j] = ((a.mat[i][k]*b.mat[k][j] % mod) + res.mat[i][j]) % mod;
            }
        }
    }
    return res;
}

data operator^ (data a , LL n) {
    data res;
    for(int i = 1 ; i <= 2 ; i++) {
        for(int j = 1 ; j <= 2 ; j++) {
            res.mat[i][j] = (i == j);
        }
    }
    while(n) {
        if(n & 1)
            res = a * res;
        a = a * a;
        n >>= 1;
    }
    return res;
}

int main()
{
    data res;
    LL n;
    cin >> res.mat[2][1] >> res.mat[1][1] >> n;
    if(n == 1) {
        cout << (res.mat[2][1] % mod + mod) % mod << endl;
    }
    else if(n == 2) {
        cout << (res.mat[1][1] % mod + mod) % mod << endl;
    }
    else {
        data a;
        a.mat[2][1] = a.mat[1][1] = 1;
        a.mat[2][2] = 0;
        a.mat[1][2] = -1;
        a = a ^ (n - 2);
        res = a * res;
        cout << (res.mat[1][1] % mod + mod) % mod << endl;
    }
}