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 · 171 阅读

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本文介绍了一种解决特定数列求值问题的方法，通过发现数列的周期性，有效地减少了计算复杂度，避免了内存溢出的问题。该方法首先计算数列的周期，然后利用周期性快速找到目标项的值。

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

代码：

#include <stdio.h>
int main(int argc,char *argv[])
{
	int c[10000];
	int a,b,n,i,t;
	c[1]=1;
	c[2]=1;
	while(scanf("%d%d%d",&a,&b,&n)&&(a!=0||b!=0||n!=0))
	{
		for(i=3;i<=1000;i++)
		{
			c[i]=(a*c[i-1]+b*c[i-2])%7;
			if(c[i]==1&&c[i-1]==1)//判断是否达到一个周期
				break;
		}
		t=i-2;
		if(n%t==0)
			printf("%d\n",c[t]);
		else
			printf("%d\n",c[n%t]);
	}
	return 0;
}