Min Number

Problem Description

Now you are given one non-negative integer n in 10-base notation, it will only contain digits ('0'-'9'). You are allowed to choose 2 integers i and j, such that: i!=j, 1≤i<j≤|n|, here |n| means the length of n’s 10-base notation. Then we can swap n[i] and n[j].

For example, n=9012, we choose i=1, j=3, then we swap n[1] and n[3], then we get 1092, which is smaller than the original n.

Now you are allowed to operate at most M times, so what is the smallest number you can get after the operation(s)?

Please note that in this problem, leading zero is not allowed!

Input

The first line of the input contains an integer T (T≤100), indicating the number of test cases.

Then T cases, for any case, only 2 integers n and M (0≤n<10^1000, 0≤M≤100) in a single line.

Output

For each test case, output the minimum number we can get after no more than M operations.

Sample Input

3 9012 0 9012 1 9012 2

Sample Output

9012 1092 1029

题意：给出一个数字串，和最多允许交换的次数。求交换后的最小数字串（首位不能为零）

题解：每次找到一个最小位与当前位交换，贪心即可

#include<stdio.h>
#include<string>
#include<iostream>
using namespace std;
int p[1010];
void swap(int &a,int &b)
{
 int temp=a;
 a=b;
 b=temp;
}
int main()
{
 int test,m,i,j,pos;
 string s;
 scanf("%d",&test);
    while(test--)
 {
        cin>>s>>m;
  int len=s.length();
  for(i=0;i<s.length();i++) p[i]=s[i]-'0';
  int maxi=10;
  int cnt=0;
  if(m)
  {
   for(i=1;i<s.length();i++)
   {
    if(p[i]<maxi&&p[i]!=0) {maxi=p[i];pos=i;}
   }
     if(maxi<p[0]) {swap(p[0],p[pos]);cnt++;}
      for(i=1;i<len-1;i++)
      {
       maxi=10;
      if(cnt==m) break;
      for(j=i+1;j<len;j++)
      {
       if(p[j]<maxi) {maxi=p[j];pos=j;}
      }
      if(maxi<p[i]) {swap(p[i],p[pos]);cnt++;}
      }
  }
  for(i=0;i<len;i++) printf("%d",p[i]);
  printf("\n");
 }
 return 0;
}