超时杭电1005



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
   
   
   
   

 

Author

CHEN, Shunbao

题目简单的要死啊，一个十位数就会超时啊，所以他一定有规律啊，但就是发现不了，看来别人的博客才知道规律是49，我简直不服啊。。。。。。

AC代码：

#include<stdio.h>
#include<math.h>
int main()
{
    int n,m,d,sum[50];
    int i;
    while(scanf("%d%d%d",&n,&m,&d)!=EOF&&n){
        sum[0]=1;
        sum[1]=1;
        sum[2]=1;
        for(i=3;i<50;i++){
             sum[i]=(n*sum[i-1]+m*sum[i-2])%7;
        }
        printf("%d\n",sum[d%49]);
    }
    return 0;
}

菜的要死了，啊啊啊，继续努力，见得太少，被水题卡N次的人。。。。。