hdu-oj 1005 Number Sequence

Number Sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 103413    Accepted Submission(s): 25045

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
 
题目大意：输入Ａ，B，n  根据f(n) = (A * f(n - 1) + B * f(n - 2)) mod 7.得到f（n）。
解题思路：考虑数据较大，f(n)应该具有规律，所以应先找到f(n)的规律进行运算。
附代码：
#include <stdio.h>
int main()
{  int a,b,n,i,m;
   int f[60];
  while(scanf("%d%d%d",&a,&b,&n),a,b,n)
  {  f[1]=1;f[2]=1;
     for(i=3;i<50;i++)
     {
       f[i]= (a* f[i-1]+ b*f[i-2])%7;
        if(f[i]==1&&f[i-1]==1)
           break;
     }//找循环规律
     m=i-2;
     if(n%m==0)
         printf("%d\n",f[m]);
     else
         printf("%d\n",f[n%m]);
  }
  return 0;
}