C - Beautiful Now

Anton has a positive integer nn, however, it quite looks like a mess, so he wants to make it beautiful after kk swaps of digits. 
Let the decimal representation of nn as (x1x2⋯xm)10(x1x2⋯xm)10 satisfying that 1≤x1≤91≤x1≤9, 0≤xi≤90≤xi≤9 (2≤i≤m)(2≤i≤m), which means n=∑mi=1xi10m−in=∑i=1mxi10m−i. In each swap, Anton can select two digits xixi and xjxj (1≤i≤j≤m)(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 kk swaps?

InputThe first line contains one integer TT, indicating the number of test cases. 
Each of the following TT lines describes a test case and contains two space-separated integers nn and kk. 
1≤T≤1001≤T≤100, 1≤n,k≤1091≤n,k≤109. 
OutputFor 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

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <algorithm>

 using namespace std;

 int maxx,minn,k,len;
 int c[],sum1[],sum2[],p[];
 char ss[];

 void update()
 {
     if(c[ p[] ]==)
     {
         return;
     }
     // 用sum1记录当前排列
     for(int i=;i<=len;++i)
     {
         sum1[i]=p[i];
     }

     int kk=,s=;
     for(int i=;i<=len;++i)
     {
         s=s* + c[ p[i] ];
         if(sum1[i] != i)
         {
             for(int j=i+;j<=len;++j)
             {
                 if(sum1[j]==i)
                 {
                     swap(sum1[i],sum1[j]);
                     ++ kk;
                     // 这一序列不能在k步内实现
                     if(kk>k)
                     {
                         return;
                     }
                     break;
                 }
             }
         }
     }

     // 当前队列满足条件
     // 更新最大最小值
     maxx=max(maxx,s);
     minn=min(minn,s);
 }

 int main()
 {
     int T;
     scanf("%d",&T);
     while(T--)
     {
         memset(sum1,,sizeof(sum1));
         memset(sum2,,sizeof(sum2));
         scanf("%s %d",ss,&k);

         len=strlen(ss);

         for(int i=;i<len;++i)
         {
             // 字符转数字
             c[i+]=ss[i]-'';
             ++ sum1[ c[i+] ];
             ++ sum2[ c[i+] ];
         }

         // 剪枝
         if(k>=len-)
         {
             // 输出最小数
             // 输出第一位（特判非零）
             for(int i=;i<=;++i)
             {
                 if(sum1[i])
                 {
                     printf("%d",i);
                     -- sum1[i];
                     break;
                 }
             }

             for(int i=;i<=;++i)
             {
                 while(sum1[i])
                 {
                     printf("%d",i);
                     --sum1[i];
                 }
             }

             printf(" ");

             // 输出最大数
             for(int i=;i>=;--i)
             {
                 while(sum2[i])
                 {
                     printf("%d",i);
                     -- sum2[i];
                 }
             }
             printf("\n");
             continue;
         }

         // 未能剪枝

         // 找一个互不相等的排列（从小到大有序）
         // 以初始排列为当前数字的编号
         for(int i=;i<=len;++i)
         {
             p[i]=i;
         }
         // 初始化最大最小值
         minn=2e9;
         maxx=-;
         // 自当前排列开始更新满足条件时的最大最小值
         do
         {
             update();
         }while(next_permutation(p+,p+len+));  // 需要头文件algor
         // 九位数的全排列，有362880种
         // 最多尝试次数，为九次
         // 复杂度约为 10^7
         printf("%d %d\n",minn,maxx);
     }
     return ;
 }