Codeforces Round #249 (Div. 2) B. Pasha Maximizes(模拟)

B. Pasha Maximizes

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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 test(s)

input

1990 1

output

9190

input

300 0

output

300

input

1034 2

output

3104

input

9090000078001234 6

output

9907000008001234

题目链接：http://codeforces.com/contest/435/problem/B

题目大意：给定一个数字，相邻位上数字可交换，求最多交换k次能得到的最大数字。

解题思路：从高位往低位枚举最高位，往后找k位以内最大的数字记为ma，如果最高位上数字小于ma，ma为最高位，k减去相应次数。

代码如下：

#include <cstdio>
#include <cstring>
char s[20];
int a[20];
int main()
{
	int n,k,p=1,j;
	scanf("%s",s);
	scanf("%d",&k);
	n=strlen(s);
	for(int i=0;i<n;i++)
		a[i]=s[i]-'0';
	if(k)
	for(int i=0;i<n;i++)
	{
		int m=k,ma=0,de=i+1; 
		for(j=i+1;j<n;j++)
		{
			m--;
			if(ma<a[j])
			{
				ma=a[j];
				de=j;
			}
			if(m<=0)
				break;
		}
		int t=a[de];
		if(a[i]<t)
		{
			for(int l=de;l>=i+1;l--)
				a[l]=a[l-1];
			a[i]=t;
			k-=(de-i);
		}
		if(k==0)
		break;
	}
	for(int i=0;i<n;i++)
		printf("%d",a[i]);
	printf("\n");
	return 0;
}