HDU1005

Number Sequence

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 1724 Accepted Submission(s): 657 

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 31 2 100 0 0

 

Sample Output

25

 

数据太大线性递推会超时，因为取模所以50以内肯定会有循环

#include <stdio.h>
#include <string.h>
int s[50];
int main()
{
    int a,b,n,i;
    while(scanf("%d%d%d",&a,&b,&n),a || b || n)
    {
       int i;
       s[0]=s[1]=1;
       for(i = 2; i<50;i++)
       {
             s[i] = (a*s[i-1]+b*s[i-2])%7;
             if(s[i] ==1 && s[i-1] == 1)
             {break;}
                   
       }  
       n = n%(i-1);
       if(n == 0)
        printf("%d/n",s[i-2]);
        else
        printf("%d/n",s[n-1]);
                                        
    }    
    return 0;
}