Codeforces Round #257 (Div. 2) B. Jzzhu and Sequences

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原文链接：http://www.cnblogs.com/mthoutai/p/6973533.html

本文提供了一道名为“Jzzhu and Sequences”的编程题的解答思路及代码实现。该题要求根据给定的x、y和n计算序列fn的值并取模10^9 + 7。通过分析发现序列具有周期性规律，周期为6，据此设计了高效的算法。

Codeforces Round #257 (Div. 2) B. Jzzhu and Sequences

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

$\text{[math]}$

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

Sample test(s)

Input

2 3
3

Output

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).

思路：有规律哒周期为6，就是最后取模的时候要注意。

AC代码:

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cmath>
#include <cstring>
#include <stdlib.h>
using namespace std;
int main(){
    int x,y;
    scanf("%d",&x);
    scanf("%d",&y);
    int n;

    scanf("%d",&n);
    int f[6]={-y+x,x,y,y-x,-x,-y};
    int temp=(f[n%6]%1000000007+1000000007)%1000000007;
    printf("%d",temp);
    return 0;
}