【杭电oj】1005 - Number Sequence（找规律）

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Number Sequence

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 151858 Accepted Submission(s): 36937

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

往后找规律就行了，为了避免前两项为1后面都为0的情况的干扰，从第三第四项开始找周期。

代码如下：

#include <cstdio>
int main()
{
	int a,b,n;
	int f[10001];
	f[1] = 1;
	f[2] = 1;
	int s;		//记录周期 
	while (~scanf ("%d %d %d",&a,&b,&n) && (a || b || n))
	{
		if (n == 1 || n == 2)
		{
			printf ("1\n");
			continue;
		}
		a %= 7;
		b %= 7;
		f[3] = (a * f[2] + b * f[1]) % 7;		//先求两项（为了避免后面都为0的情况） 
		f[4] = (a * f[3] + b * f[2]) % 7;
		for (int i = 5 ; ; i++)
		{
			f[i] = (a * f[i-1] + b * f[i-2]) % 7;
			if (f[i] == f[4] && f[i-1] == f[3]) 		//两项相等就开始循环
			{
				s = i - 4;
				break;
			}
		}
//		int pos = (n - 2) % s;
//		pos = pos == 0 ? s : pos;		//余数为0取最后一项
		int pos = (n - 3) % s + 1;		//这样的写法比上面的巧，小细节，学习一下 
		printf ("%d\n",f[2+pos]);
	}
	return 0;
}