CodeForces

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+7;
const int MAXN = 1e6+100;

const int Size = 2;
struct MA
{
    LL mat[Size][Size];
    void init()
    {
        for(int i = 0; i<Size; i++)
        for(int j = 0; j<Size; j++)
            mat[i][j] = (i==j);
    }
};

MA mul(MA x, MA y)
{
    MA ret;
    memset(ret.mat, 0, sizeof(ret.mat));
    for(int i = 0; i<Size; i++)
    for(int j = 0; j<Size; j++)
    for(int k = 0; 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&1) s = mul(s, x);
        x = mul(x, x);
        y >>= 1;
    }
    return s;
}

MA tmp = {
    1, -1,
    1, 0
};

int main()
{
    LL f[3], n;
    while(scanf("%lld%lld%lld",&f[1],&f[2],&n)!=EOF)
    {
        if(n<=2)
        {
            printf("%lld
", (f[n]+MOD)%MOD);
            continue;
        }

        MA s = tmp;
        s = qpow(s, n-2);
        LL ans = (1LL*s.mat[0][0]*f[2]%MOD+1LL*s.mat[0][1]*f[1]%MOD+2*MOD)%MOD;
        printf("%lld
", ans);
    }
}