Codeforces 450 B. Jzzhu and Sequences

最新推荐文章于 2019-09-25 20:45:00 发布

转载最新推荐文章于 2019-09-25 20:45:00 发布 · 49 阅读

本文探讨了一种特殊序列的模运算计算方法，包括输入参数x、y及序列索引n，通过简化计算步骤，实现了快速求解fn模1000000007的结果。详细解释了序列特性及其应用实例，并提供了优化后的代码实现，旨在提升复杂计算任务的效率。

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6单周期.....模时间很坑...

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

$\text{[math]}$

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).

Sample test(s)

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).

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

typedef long long int LL;

const LL mod=1000000007LL;
LL a,b,n;

int main()
{
    cin>>a>>b>>n;
    n--;
    n=n%6;
    LL ans=0;
    switch(n)
    {
        case(0): 
        {
            ans=(a+mod)%mod;
            break;
        }
        case(1):
        {
            ans=(b+mod)%mod;
            break;
        }
        case(2):
        {
            ans=(b-a+2*mod)%mod;
            break;
        }
        case(3):
        {
            ans=(mod-a)%mod;
            break;
        }
        case(4):
        {
            ans=(mod-b)%mod;
            break;
        }
        case(5):
        {
            ans=(a-b+2*mod)%mod;
            break;
        }
    }
    cout<<ans<<endl;
    return 0;
}