Codeforces Round #257 (Div. 2) B. Jzzhu and Sequences-优快云博客

本文链接：https://blog.youkuaiyun.com/Kimi_R_17/article/details/38059609

本文提供了一种解决 B.Jzzhu 所发明的特定序列问题的方法，包括输入解析、算法实现和输出展示。重点在于通过周期性的规律来计算序列的指定项并进行模运算。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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;
}