Number Sequence hdu1005 数学问题

题意：

Number Sequence

Problem Description 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

原码加思路：

/*
做题方法：
 前是打表，就是打出一大堆数据，然后发现规率：
发现这道题是从f[1]=1,和f[2]=1开始，然后依次模7，就可知f[n]只有7种情况，而相数相邻只有7*7=49种
，所以从f[1]到f[49]必会出现相邻两个f[m-1]=1,f[m]=1，所以f[m]为周期函数，49为其一个周期;
代码如下：*/
#include<stdio.h>
int main()
{
    int f[56],a,b,i;
    f[0]=1;f[1]=1;
    int n;
    while(1)
    {
        scanf("%d%d%d",&a,&b,&n);
        if(!a && !b && !n)
            break;
        for(i=2;i<49;i++)
            f[i]=(a*f[i-1]+b*f[i-2])%7;
        printf("%d\n",f[(n-1)%49]);
    }
    return 0;
}