B. Secret Combination

最新推荐文章于 2018-10-04 10:15:07 发布

原创最新推荐文章于 2018-10-04 10:15:07 发布 · 576 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#codeforces

codeforces 专栏收录该内容

32 篇文章

订阅专栏

本文介绍了一种通过操作组合锁的按钮来找到显示面板上能够构成的最小数字的方法。面对一个具有特定初始状态的n位数字显示面板，可以通过两种操作（增加每位数字或循环右移数字）来改变显示的数字。目标是找到经由这两种操作后所能得到的最小可能数字。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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.

Sample test(s)

input
3
579

output
024

input
4
2014

output
0142这道题可以用暴力过，每次把最前的一个字符移到最后，然后把第一个字符变成'0'，后面依次变化，求出最小的字符串。#include<stdio.h>
#include<string.h>
char s[1005],str[1005],str1[1005];
char word[40]={'0','1','2','3','4','5','6','7','8','9','0','1','2','3','4','5','6','7','8','9','0','1','2','3','4','5','6','7','8','9','0'};
int main()
{
	int n,m,i,j,len,t;
	char c;
	while(scanf("%d",&n)!=EOF)
	{
		memset(s,0,sizeof(s));
		scanf("%s",s);
		len=strlen(s);
		for(i=1;i<=len;i++){
			if(i!=1){
				c=s[0];
				for(j=1;j<=len-1;j++){
					s[j-1]=s[j];
				}
				s[len-1]=c;
			}
			strcpy(str,s);
			t='9'-s[0]+1;
			str[0]='0';
			for(j=1;j<len;j++){
				str[j]=word[str[j]-'0'+t];
			}
			str[len]='\0';
			if(i==1 || strcmp(str1,str)>0){
				strcpy(str1,str);
			}
			
		}
		printf("%s\n",str1);
	}
	return 0;
}