找规律的题

Number Sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 41692    Accepted Submission(s): 8995

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

 

Author

CHEN, Shunbao

 

Source

ZJCPC2004

 

Recommend

JGShining

 

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自己写了一个找  周期 但是就是ac不了，下面的是百度上找的

#include<stdio.h>
int main()
{     int a,b,n;    
 while(scanf("%d %d %d",&a,&b,&n)!=EOF&&(a+b+n)!=0)
 {  
 int f1=1,f2=1,f3=1,t,r[8][8]={0},i,flag=0,x;        
  for(i=3;i<=n;i++)
  {      f3=(a*f2+b*f1)%7;       
        r[f1][f2]=i;            
   f1=f2;             
   f2=f3;            
    if(r[f1][f2]&&flag==0)
    {               
      t=i-r[f1][f2]+1;               
        i=r[f1][f2]-3;       
      
   n=(n-i)%t+i  ;        
      if(n==i)                
        n=t+i;                 
     if(n==(i+1)) {f3=f1;break;}   
          if(n==(i+2)) {f3=f2;break;}              
          i+=2;          
         flag=1;             
       }   
           
  }         
     printf("%d\n",f3);     
  }     
  return 0;
}