Number Sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 182573    Accepted Submission(s): 45387


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 感覺程式碼有點不妥，但是能過，。。。。。。。 程式碼：

#include<iostream>
#include<algorithm>
#include<cstring>
#include<cstdio>
int s[100];
int main()
{
    int a,b,n;
    while(scanf("%d%d%d",&a,&b,&n)&&a+b+n)
    {
        int count=0;
        if(a%7==0&&b%7==0)
        {
          if(n==1||n==2)
          {
             printf("1\n");
          }
          else
          {
            printf("0\n");
          }
          count=1;
        }
        if(count)
         continue;
        s[1]=s[2]=1;
        for(int i=3;i<50;i++)
        {
            s[i]=(a*s[i-1]+b*s[i-2])%7;
        }
        printf("%d\n",s[n%48]);

    }
    return 0;

}