1%x,就是1;
x%x=0;
開數組a[x]不是a[x],是a[0];
http://codeforces.com/contest/450/problem/B
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).
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 a single integer representing fn modulo 1000000007 (109 + 7).
2 3 3
1
0 -1 2
1000000006
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).
最後處理注意下
#include<iostream>
using namespace std;
int main()
{
long long n;
long long a[100];
cin>>a[1]>>a[2];
cin>>n;
a[3]=a[2]-a[1];
a[4]=-a[1];
a[5]=-a[2];
a[0]=-a[3];
int ss = n % 6;
ss = a[ss];
ss %= 1000000007;
if (ss < 0)
ss += 1000000007;
printf("%d\n", ss);
}