【杭电1005】Number Sequence

Number Sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 158133    Accepted Submission(s): 38729

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   |   We have carefully selected several similar problems for you:   1008   1004   1021   1019   1009  

可以发现一个规律：因为对7取模，f[n]不超过7，从中间或者开头开始形成一个周期，找到这个周期循环就可以

code:

<span style="font-size:14px;">#include<cstdio>
long long f[100000000];
int main()
{
	long long n;
	int a,b,j;
	while(scanf("%d%d%lld",&a,&b,&n)!=EOF&&a!=0&&b!=0&&n!=0)
	{
		f[1]=1;f[2]=1;
		int s=0;//记录周期 
		for(int i=3;i<=n;i++)
		{
		f[i]=(a*f[i-1]+b*f[i-2])%7;	
		for(j=2;j<i;j++)
		{
			if(f[i]==f[j]&&f[i-1]==f[j-1])//确定前两个找到一个周期
			{s=i-j;break;
			} 
		}
		if(s>0)break;
		}
		if(s>0)
		f[n]=f[(n-j)%s+j];//周期循环并加上前面不参加循环的数
		printf("%lld\n",f[n]);
	}
	return 0;
 } </span>

另一种code：

#include<iostream>  
using namespace std;  
int f(int A, int B,int n)  
{  
    if(n==1 || n==2)  
        return 1;  
    else  
        return (A*f(A,B,n-1)+B*f(A,B,n-2))%7;  
}  
int main()  
{  
    int a,b,n;  
    while(cin>>a>>b>>n,a||b||n)  
    {  
        cout<<f(a,b,n%49)<<endl;  
    }  
    return 1;  
}