hdu1005(大数计算可能有规律)

1)原题代码

#include<stdio.h>
int main()
{
    int a,b,c,i; 
   int ab[50]; 
   ab[1]=1;ab[2]=1;
    while(scanf("%d%d%d",&a,&b,&c))  
  {    
    if(a==0&&b==0&&c==0)
break;  
      for(i=3;i<50;i++)  
      {     
       ab[i]=(a*ab[i-1]+b*ab[i-2])%7;    
    }   
     printf("%d\n",ab[c%49]);
    } 
   return 0;  
  }

2）

用递归因为调用栈帧过多会溢出，用循环因为计算次数过大会超时。

写前几个结果发现有规律，f(n-1)，f(n-2)都在0~6这7个数的范围内，而A、B都是常数，所以A*f(n-1)+B*f(n-2)有7*7共49种组合，所以只记录前50个结果即可，因此建立啊a[50]，并用循环计算。

3）原题

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