CodeForces - 450B Jzzhu and Sequences —— 斐波那契数、矩阵快速幂

题目链接：https://vjudge.net/problem/CodeForces-450B

B. Jzzhu and Sequences

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Jzzhu has invented a kind of sequences, they meet the following property:

You are given x and y, please calculate fn modulo 1000000007 (109 + 7).

Input

The first line contains two integers x and y (|x|, |y| ≤ 109). The second line contains a single integer n (1 ≤ n ≤ 2·109).

Output

Output a single integer representing fn modulo 1000000007 (109 + 7).

Examples

input

2 3
3

output

1

input

0 -1
2

output

1000000006

Note

In the first sample, f2 = f1 + f3, 3 = 2 + f3, f3 = 1.

In the second sample, f2 =  - 1;  - 1 modulo (109 + 7) equals (109 + 6).

题解：

斐波那契数，递推式为：f[n] = f[n-1] - f[n-2]，构造矩阵如下：

代码如下：

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <algorithm>
 #include <vector>
 #include <cmath>
 #include <queue>
 #include <stack>
 #include <map>
 #include <string>
 #include <set>
 using namespace std;
 typedef long long LL;
 const int INF = 2e9;
 const LL LNF = 9e18;
 const int MOD = 1e9+;
 const int MAXN = 1e6+;

 const int Size = ;
 struct MA
 {
     LL mat[Size][Size];
     void init()
     {
         for(int i = ; i<Size; i++)
         for(int j = ; j<Size; j++)
             mat[i][j] = (i==j);
     }
 };

 MA mul(MA x, MA y)
 {
     MA ret;
     memset(ret.mat, , sizeof(ret.mat));
     for(int i = ; i<Size; i++)
     for(int j = ; j<Size; j++)
     for(int k = ; k<Size; k++)
         ret.mat[i][j] += (1LL*x.mat[i][k]*y.mat[k][j])%MOD, ret.mat[i][j] %= MOD;
     return ret;
 }

 MA qpow(MA x, LL y)
 {
     MA s;
     s.init();
     while(y)
     {
         if(y&) s = mul(s, x);
         x = mul(x, x);
         y >>= ;
     }
     return s;
 }

 MA tmp = {
     , -,
     ,
 };

 int main()
 {
     LL f[], n;
     while(scanf("%lld%lld%lld",&f[],&f[],&n)!=EOF)
     {
         if(n<=)
         {
             printf("%lld\n", (f[n]+MOD)%MOD);
             continue;
         }

         MA s = tmp;
         s = qpow(s, n-);
         LL ans = (1LL*s.mat[][]*f[]%MOD+1LL*s.mat[][]*f[]%MOD+*MOD)%MOD;
         printf("%lld\n", ans);
     }
 }