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

 1 #include <stdio.h>
 2 #include <stdlib.h>
 3 #include <iostream>
 4 #include <algorithm>
 5 #include <math.h>
 6 
 7 int main()
 8 {
 9     long long   n,m,i,j,k,t;
10     long long     str[7];
11     while(scanf("%I64d%I64d",&n,&m)!=EOF)
12     {
13         scanf("%I64d",&t);
14         str[0]=(n+1000000007)%1000000007;
15         str[1]=(m+1000000007)%1000000007;
16         for(i=2;i<6;i++)
17         {
18             str[i]=(str[i-1]-str[i-2]);//printf("%I64d*",t%6);
19         }
20         printf("%I64d\n",(str[(t-1)%6]+1000000007)%1000000007);
21         //else
22         //printf("%I64d\n",str[t%6]%1000000007);
23     }
24     return 0;
25 }