FZU 2111【 Min Number】

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



用原串与最小的排列顺序进行比较。

一位一位的比较就行了。不一样就换一次、

#include <iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
using namespace std;
char s[1110],s1[1110],s2[1110];
int main()
{
int t,n;
char tmp;
scanf("%d",&t);
while(t--)
{
memset(s,0,sizeof(s));
memset(s1,0,sizeof(s1));

scanf("%s%d",s,&n);
strcpy(s1,s);//复制
int len=strlen(s1);
sort(s1,s1+len);
if(s1[0]=='0')//第一个位置不为0
{
for(int i=0; i<len; i++)
{
if(s1[i]!='0')
{
tmp=s1[i];
s1[i]=s1[0];
s1[0]=tmp;
break;

}
}
}
//此时是s1是最小排序

while(n--)
{
int flag=0;
for(int i=0; i<len; i++)
{
if(s[i]!=s1[i])//逐位扫描
{
for(int j=i; j<len; j++)
{
if(s[j]==s1[i])
{
tmp=s[j];
s[j]=s[i];
s[i]=tmp;
flag=1;
break;
}
}
}
if(flag)
break;
}
}

for(int i=0; i<len; i++)
printf("%c",s[i]);
cout<<endl;

}

return 0;
}