Number Sequence

最新推荐文章于 2024-08-07 14:16:33 发布

原创最新推荐文章于 2024-08-07 14:16:33 发布 · 183 阅读

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本文探讨了一种特定的周期性数列计算方法，通过分析数列的周期性特征，提出了一个高效的算法来计算数列中任意一项的值。该算法适用于给定参数A、B和n的情况下，快速计算数列f(n)的值，其中f(n)=(A*f(n-1)+B*f(n-2))mod7。通过对周期性的利用，避免了传统递归或迭代方法的效率瓶颈。

Problem
DescriptionA 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

#include<stdio.h>
int Fn( int a,int b,int n )
{
 	int  i=1,j=1,m;
 	int x=1,y=1;
 	if(n==1) return 1;
 	else if(n==2)  return 1;
 	else 
 	{
  		n=(n-2)%16; //判断周期(周期16) 
  		while(n--)
 	 	{
   			x=((x*b)%7+(y*a)%7)%7;
   			m=x;x=y;y=m;
  		}
  	return m;		
	} 
}
int main()
{
 	int a,b,n;
	while(scanf("%d %d %d",&a,&b,&n)&&(a!=0&&b!=0&&n!=0))
  		printf("%d\n",Fn(a,b,n));
 	return 0;
}