Number Sequence hdu1005

Number Sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 37475    Accepted Submission(s): 7887

Problem Description

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

 

 

Source

ZJCPC2004

 

 

Recommend

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)I]);

    }

    return 0;

}