HDU 1005 Number Sequence

最新推荐文章于 2024-07-18 12:00:00 发布

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最新推荐文章于 2024-07-18 12:00:00 发布

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Number Sequence

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 188189 Accepted Submission(s): 46920

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

直接写肯定超时要不就爆内存，要优雅的暴力，不能无脑暴力。

这种题要找到他的循环节然后去找答案。

#include<iostream>
#include<cstring>
#include<string>
#include<algorithm>
#include<map>
using namespace std;
int res[55];
int main()
{
	int a, b, n;
	res[1] = res[2] = 1;
	while (cin >> a >> b >> n&&a&&b&&n)
	{
		bool flag = 0;
		int resl, resr;
		for (int i = 3; i <= n&&!flag; i++)
		{
			res[i] = (a*res[i - 1] + b*res[i - 2]) % 7;
			for (int j = 2; j < i; j++)
			{
				if (res[i] == res[j] && res[i - 1] == res[j - 1])
				{
					resl = j;
					resr = i;
					flag = 1;
					break;
				}
			}
		}
		if (flag)
			cout << res[resl + (n - resr) % (resr - resl)] << endl;
		else
			cout << res[n] << endl;
	}
	return 0;
}