Beautiful Now

Time Limit: 5000/2500 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 2386 Accepted Submission(s): 866

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

Source

2018 Multi-University Training Contest 5

看了好多博客，发现有用搜索的，有用全排列的，在这里全排列是可以暴力过的

最大的难度是次数怎样去实现，用下标来实现，当访问过的就标记为1，没访问过的标记为0，显然访问过的不能再进行下去，跳出循环，这样去卡它的次数

用全排列的好处是，它会自动排好序，例如

#include <stdio.h>
#include <algorithm>
using namespace std;
int main(){
    int n;
    while(scanf("%d",&n)&&n){
        int a[1000];
        for(int i=0;i<n;i++){
            scanf("%d",&a[i]);
        }
        sort(a,a+n);//可以自行测试一下删除后的结果
        do{
            for(int i=0;i<n;i++)
                printf("%d ",a[i]);
            printf("\n");
        }while(next_permutation(a,a+n));
    }
    return 0;
}

3
7 5 3
3 5 7
3 7 5
5 3 7
5 7 3
7 3 5
7 5 3

只需判断出符合次数条件，且在此基础上还是最大的，最小的即可

全排列下标，因为字符可以重复，若不用此方法，计算次数的时候可能会WA

#include<bits/stdc++.h>
using namespace std;
int vis[20];
int xb[20];//下标
int k,len;
char s[20];//字符数组

int cheak()//检查交换次数
{
    fill(vis,vis+len,0);每个字符长度不一样，所以用fill会将不同长度的字符数组置为0
    int p=0;
    for(int i=0;i<len;i++)
    {
        if(vis[i])  continue;
        int ans=0;
        while(vis[i]==0)
        {
            ans++;
            vis[i]=1;
            i=xb[i];
        }
         p=p+ans-1;
         if(p>k)  return 0;//可能在小于k的时间就已经找到最小和最大值
     }
     return 1;
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%s %d",s,&k);
        len=strlen(s);
        for(int i=0;i<len;i++)
        {
            xb[i]=i;//下标初始化
        }
        int max1=0;
        int min1=1e9+5;
        do
        {

          if(s[xb[0]]!='0'&&cheak())
          {
              int sum=0;
              for(int i=0;i<len;i++)
              {
                  sum=sum*10+s[xb[i]]-'0';
              }
              if(sum<min1)
              {
                  min1=sum;
              }
              if(sum>max1)
              {
                  max1=sum;
              }
          }
        }while(next_permutation(xb,xb+len));//全排列
        printf("%d %d\n",min1,max1);
    }
    return 0;

}

怎样去查找它的次数

          i       0 1 2 3
数组a[i]          1 4 7 5
下标b[i]          0 1 2 3

          i       0 1 2 3
交换一次 下标b[i] 0 1 3 2
             a[i] 1 4 5 7
可以发现如果下标2，3改变，则去看当i=2的时候，下标对应的是b[2]=3,
将b[2]的值作为下标去检查，此时当i=3时看是否b[3]与原序列改变
这个规律之前我也没发现，真的很神奇。。。。

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