HDU 1005Number Sequence(循环节)

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

ZJCPC2004

感觉代码有点不妥，但是能过，。。。。。。。

代码：

#include
#include
#include
#include
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;

}