HDU1005

A number sequence is defined as follows: 

f(1) = 1, f(2) = 1, f(n) = (A * f(n - 1) + B * f(n - 2)) mod 7. 

Given A, B, and n, you are to calculate the value of f(n). 

Input

The input consists of multiple test cases. Each test case contains 3 integers A, B and n on a single line (1 <= A, B <= 1000, 1 <= n <= 100,000,000). Three zeros signal the end of input and this test case is not to be processed. 

Output

For each test case, print the value of f(n) on a single line. 

Sample Input

1 1 3
1 2 10
0 0 0

Sample Output

2
5

思路：常见的解法是矩阵快速幂，我还不会-_-||，而用递归来写的话则会超时，需要进行优化

由同余定理：(a+b)%c=(a%c+b%c)%c     ,而题中要求%7，则a和b取值在0到6之间，所以共有7*7=49种情况，往后会成循环了，所以要将题中m改为m%49，即可AC；

#include <iostream>

using namespace std;
int fb(int x,int y,int n)
{
    if(n==1||n==2)
        return 1;
    return (x*fb(x,y,n-1)+y*fb(x,y,n-2))%7;
}
int main()
{
    int x,y,m;
    while(cin>>x>>y>>m && x+y+m)
    {
        int a=fb(x,y,m%49);
        cout<<a<<endl;
    }
    return 0;
}