CodeForces_435B_C

Time Limit:1000MS Memory Limit:262144KB 64bit IO Format:%I64d
& %I64u
Submit Status Practice CodeForces
435B

Description

Pasha has a positive integer a without leading zeroes. Today he decided that the number is too small and he should make it larger. Unfortunately, the only operation Pasha can do is to swap two
adjacent decimal digits of the integer.

Help Pasha count the maximum number he can get if he has the time to make at most k swaps.

Input

The single line contains two integers a and k(1 ≤ a ≤ 1018; 0 ≤ k ≤ 100).

Output

Print the maximum number that Pasha can get if he makes at most k swaps.

Sample Input

Input

1990 1

Output

9190

Input

300 0

Output

300

Input

1034 2

Output

3104

Input

9090000078001234 6

Output

9907000008001234

这道题是贪心的思想，在n步之内找到最大的数字，从后往前交换，交换一次n--；直到n为零；

#include <iostream>

#include <stdio.h>

#include <string.h>

char a[20];

int n;

using namespace std;

int main()

{

while(scanf("%s",&a)!=EOF)

{

scanf("%d",&n);

int len=strlen(a);

int npos,j;

for(int i=0;i<len;i++)

{

npos=i;

char ch=a[i];

for( j=i+1;j<=i+n&&j<len;j++)

{

if(a[j]>ch)

{

npos=j;

ch=a[j];

}

}

//cout<<npos;

if(npos!=i)

{

for(int p=npos;p>=i+1;p--)

{

char tem=a[p];

a[p]=a[p-1];

a[p-1]=tem;

n--;

}

}

if(!n) break;

}

for(int i=0;i<len;i++) cout<<a[i];

printf("\n");

}

return 0;

}