[hdoj_1005]Number Sequence

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1005

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

题目大意：

1.多组测试数据，依次输入A,B,n(1 <= A, B <= 1000, 1 <= n <= 100,000,000)，输入0 0 0 终止；

2.f(1) = 1, f(2) = 1, f(n) = (A * f(n - 1) + B * f(n - 2)) mod 7；

3.输出 f(n)。

解题思路：

1.n的数值很大，这类数值很大的问题一般都有规律，找出循环节（周期）是关键；

2.找规律，这道题是从 f(1) = 1 和 f(2) = 1 开始，然后依次模7，可知 f(n) 只有7种情况，所以两数相邻只有7*7=49种；

3.所以从 f(1) 到 f(49) 必会出现相邻两个 f(m-1) = 1 , f(m) = 1，所以 f(n) 为周期函数，49为其一个周期；

需要注意的地方：

做题的时候不要盲目的上手去做，先要仔细分析题目，先动脑再动手。

代码如下：

// hdoj_1005 Number Sequence
// 31MS	220K	370 B	GCC

#include 

int fun(int a, int b, __int64 n);

int main(void)
{
	int a, b;
	__int64 n;
	while(scanf("%d%d%I64d", &a, &b, &n) && a != 0 && b != 0 && n != 0)
	{
		printf("%d\n", fun(a, b, n%49));
	}
	return 0;
}

int fun(int a, int b, __int64 n)
{
	if(n == 1 || n == 2)
		return 1;
	else
		return (a*fun(a, b, n-1) + b*fun(a, b, n-2)) % 7;
}