Codeforces Round #257 (Div. 2) B

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).

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).

 ----------------------------------------------------------------------------

题意就不说了，找最小循环节点，就是6

然后注意数组要从0开始，并且负数取余要让它加上一个整数（1000000007）然后再取余，就没什么了

#include <stdio.h>
#include <stdlib.h>
#include <iostream>
#include <algorithm>
#include <math.h>

int main()
{
    long long   n,m,i,j,k,t;
    long long     str[7];
    while(scanf("%I64d%I64d",&n,&m)!=EOF)
    {
        scanf("%I64d",&t);
        str[0]=(n+1000000007)%1000000007;
        str[1]=(m+1000000007)%1000000007;
        for(i=2;i<6;i++)
        {
            str[i]=(str[i-1]-str[i-2]);//printf("%I64d*",t%6);
        }
        printf("%I64d
",(str[(t-1)%6]+1000000007)%1000000007);
        //else
        //printf("%I64d
",str[t%6]%1000000007);
    }
    return 0;
}