HDUOJ1005Number Sequence

Number Sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 61101    Accepted Submission(s): 13923

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

 

#include<stdio.h>
#include<string.h>
int str[200];
int main()
{
    int i,n,a,b,k;
    while(scanf("%d %d %d",&a,&b,&n)&&(a||b||n))
    {
        memset(str,0,sizeof(str));
        str[0]=1,str[1]=1;
        for(i=2;i<200;i++)
        {
            str[i]=(a*str[i-1]+b*str[i-2])%7;
            //printf("str[%d]=%d",i,str[i]);//
            if(str[i]==1&&str[i]==str[i-1])
            {   //k=i;
               //printf("i=%d k==%d",i,k);
                  break;
            }
        }
        printf("%d\n",str[(n-1)%(i-1)]);
    }
    return 0;
}