Codeforces Round #283 (Div. 2)——B. Secret Combination

最新推荐文章于 2022-12-16 16:19:21 发布

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本文介绍了一个关于组合锁最小数字开锁的问题。通过将循环展开为链，并利用最多加9次的限制，实现暴力求解。该方法通过复制、循环移位和比较数字来找出最小可能的数字状态，最终输出解。代码示例展示了如何通过简单的循环和条件判断，高效地解决实际问题。

You got a box with a combination lock. The lock has a display showing n digits. There are two buttons on the box, each button changes digits on the display. You have quickly discovered that the first button adds 1 to all the digits (all digits 9 become digits 0), and the second button shifts all the digits on the display one position to the right (the last digit becomes the first one). For example, if the display is currently showing number 579, then if we push the first button, the display will show 680, and if after that we push the second button, the display will show 068.

You know that the lock will open if the display is showing the smallest possible number that can be obtained by pushing the buttons in some order. The leading zeros are ignored while comparing numbers. Now your task is to find the desired number.

Input

The first line contains a single integer n (1 ≤ n ≤ 1000) — the number of digits on the display.

The second line contains n digits — the initial state of the display.

Output

Print a single line containing n digits — the desired state of the display containing the smallest possible number.

水题，把环展开成链，而且最多加9次，所以暴力就行

#include <map>
#include <set>
#include <list>
#include <queue>
#include <stack>
#include <vector>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

char str[3000];
char ans[3000];
int main()
{
	int n;
	while (~scanf("%d", &n))
	{
		scanf("%s", str);
		for (int i = n; i < 2 * n - 1; ++i)
		{
			str[i] = str[i - n];
		}
		str[2 * n - 1] = '\0';
		for (int i = 0; i < n; ++i)
		{
			ans[i] = '9';
		}
		ans[n] = '\0';
		int ret = 1;
		while (ret <= 10)
		{
			for (int i = 0; i < n; ++i)
			{
				int cnt = 0;
				bool flag = false;
				while (ans[cnt] == '0')
				{
					cnt++;
				}
				int j = i;
				while (str[j] == '0')
				{
					j++;
				}
				if (n - cnt > i + n - j)
				{
					flag = true;
				}
				else if (n - cnt == i + n - j)
				{
					for (int k = j; k < i + n; ++k)
					{
						if (ans[cnt] == str[k])
						{
							cnt++;
						}
						else if (ans[cnt] > str[k])
						{
							flag = true;
							break;
						}
						else
						{
							break;
						}
					}
				}
				if (flag)
				{
					cnt = 0;
					for (int k = i; k < i + n; ++k)
					{
						ans[cnt++] = str[k];
					}
				}
				ans[n] = '\0';
			}
			for (int i = 0; i < 2 * n - 1; ++i)
			{
				int tmp = str[i] - '0';
				tmp = (tmp + 1) % 10;
				str[i] = tmp + '0';
			}
			str[2 * n - 1] = '\0';
			ret++;
		}
		printf("%s\n", ans);
	}
	return 0;
}