hdu1005 Number Sequence(数论)

最新推荐文章于 2020-02-07 11:05:28 发布

LittleLoveBoy

最新推荐文章于 2020-02-07 11:05:28 发布

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分类专栏：数论

本文链接：https://blog.youkuaiyun.com/u011361299/article/details/12655187

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本文探讨了一个特定数列的定义及其求解方法，通过实例展示了如何利用周期性规律来简化计算过程，特别关注了参数A、B与序列长度n之间的相互作用，以及在实际输入数据下如何高效地计算数列的第n项。

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Number Sequence

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 86102 Accepted Submission(s): 20423

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

Sample Output

2
5

Author

CHEN, Shunbao

Source

ZJCPC2004

当时别人问我这题，我最开始一看感觉是矩阵快速幂，回来后想想%7，因为7太小了，可能有规律，周期最大为48，突然想起来原来以前做过这个题

#include <stdio.h>
int main()
{
	int i,j,A,B,f1,f2,f3,n;
	while(scanf("%d%d%d",&A,&B,&n)!=EOF&&(A||B||n))
	{
		f1=1;
		f2=1;
		A=A%7;
		B=B%7;
		n=(n-3)%48+2;
		for(i=2;i<=n;i++)
		{
			f3=(A*f2+B*f1)%7;
			f1=f2;
			f2=f3;
		}
		printf("%d\n",f2);
	}
	return 0;
}