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 + ;
 struct data {
     LL mat[][];
 };

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

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