Problem Description

Anton has a positive integer n , however, it quite looks like a mess, so he wants to make it beautiful after k swaps of digits.
Let the decimal representation of n as (x1x2⋯xm)10 satisfying that 1≤x1≤9 , 0≤xi≤9 (2≤i≤m) , which means n=∑mi=1xi10m−i . In each swap, Anton can select two digits xi and xj (1≤i≤j≤m) and then swap them if the integer after this swap has no leading zero.
Could you please tell him the minimum integer and the maximum integer he can obtain after k swaps?

Input

The first line contains one integer T , indicating the number of test cases.
Each of the following T lines describes a test case and contains two space-separated integers n and k .
1≤T≤100 , 1≤n,k≤109 .

Output

For each test case, print in one line the minimum integer and the maximum integer which are separated by one space.

Sample Input

5 12 1 213 2 998244353 1 998244353 2 998244353 3

Sample Output

12 21 123 321 298944353 998544323 238944359 998544332 233944859 998544332

這份程式碼是參考後發現的最容易理解且最簡單的程式碼，思路明確。

#include<stdio.h>
#include<string>
#include<iostream>
#include<algorithm>
#include<string.h>
using namespace std;
int Hash[20];
int vis[20]; 
int n,k;
bool check() //檢查交換次數
{
	fill(vis,vis+n,0);
	int sw=0;
	for(int i=0;i<n;i++)
	{
		if(vis[i])continue;
		int x=0;
		while(vis[i]==0)
		{
			x++;
			vis[i]=1;
			i=Hash[i];
		}
		sw+=x-1;
		if(sw>k)
		return false;//為了不超時這裡不符合條件的要提前退出
	}
	return true;
}
int main()
{
	char s[20];
	int t;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%s%d",s,&k);
		n=strlen(s);
		
		for(int i=0;i<n;i++)
		{
			Hash[i]=i;
		}
	
		int Min=1e9+7;
		int Max=0;
		do{
			if(s[Hash[0]]!='0'&&check())//不能有前導0，且交換次數等於k; 
			{	
				int sum=0;
				for(int i=0;i<n;i++)
				{
					sum=sum*10+s[Hash[i]]-'0';
				}
				if(sum<Min)
				{
					Min=sum;
				}
				if(sum>Max)
				{
					Max=sum;
				}
			}
		}while(next_permutation(Hash,Hash+n));//全排列
		printf("%d %d\n",Min,Max);
	}
 }