Number Sequence

Number Sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 43278    Accepted Submission(s): 9449

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

 

这个题目的规律性在于有一个mod存在，这个也是数值存在规律的前提。可经过验证满足没49个循环一遍，因此程序为：

#include<stdio.h>

int main()
{
 int a1[50];
 int i,a,b,c;
 while(scanf("%d%d%d",&a,&b,&c)!=EOF && (a!=0 || b!=0 || c!=0))
 {
  a=a%7;
  b=b%7;
  a1[0]=1;
  a1[1]=1;
  for(i=2;i<48;i++)
   a1[i]=(a1[i-2]*b+a1[i-1]*a)%7;
  c%=48;
  printf("%d\n",a1[c]);
 }
 return 0;
}