HDU 1005.Number Sequence【很多问题是不能直接求的】【8月15】

Number Sequence

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
   
   
   
   
   
   
   
   


   
   
   
   

f[1]=f[2]=1;f[n]=(A*f[n-1]+B*f[n-2])%7;给定A，B，n，求f[n]。

不能直接求，不然肯定超时超内存。寻找循环。f[1]=1,f[2]=1;当连续出现两个1的时候就说明是循环了。代码如下：

#include<cstdio>
int main(){
    int a,b,n,f[11000];
    f[1]=1;f[2]=1;
    while(scanf("%d%d%d",&a,&b,&n)!=EOF,a||b||n){
        int i;
        for(i=3;i<11000;i++){
            f[i]=(a*f[i-1]+b*f[i-2])%7;
            if(f[i]==1&&f[i-1]==1)//i-2就是循环长度
                break;
        }
        f[0]=f[i-2];
        printf("%d\n",f[n%(i-2)]);
    }
    return 0;
}