HDOJ 1005 Number Sequence规律题

本来想每次模7就够了，但这样忽略了数据量的庞大，规律如下：

因为模总是在0~6之间，所以f(i-1)与f(i-2)也是如此，每种有七种情况，所以公式就有7X7=49种情况，49为一循环

        代码如下：

#include<stdio.h>
int f[100000005];
int main()
{
    f[1]=1;
    f[2]=1;
    int i,p,q,n;
    while(~scanf("%d%d%d",&p,&q,&n)&&(p||q||n))
    {
        for(i=3;i<=49;i++)
        {
            f[i]=(p*f[i-1]+q*f[i-2])%7;
        }
        printf("%d\n",f[n%49]);
    }
}

题目如下：

Number Sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 140179    Accepted Submission(s): 33985

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