FZOJ2111：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n和交换次数k，求对这个数进行k次交换后得到的最小的数是什么

思路：对这个数两头进行枚举即可

 

#include <stdio.h>
#include <string.h>

int main()
{
	int n,i,j,len,l,MIN,flag,ss;
	char str[1005],min_c,t;
	scanf("%d",&ss);
	while(ss--)
	{
		scanf("%s%d",str,&n);
		if(!n)
		{
			printf("%s\n",str);
			continue;
		}
		len = strlen(str);
		l = 0;
		min_c = str[0];
		for(i = len-1;i>0;i--)//找出整个串最小的，且不为0的放到第一位
		{
			if(str[i]!='0' && str[i]<min_c)
			{
				min_c = str[i];
				flag = i;
			}
		}
		if(min_c!=str[0])//交换
		{
			t = str[flag];
			str[flag] = str[0];
			str[0] = t;
			n--;
		}
		for(i = 1;i<len;i++)//从第二位开始
		{
			if(!n)
				break;
			min_c = str[i];
			for(j = len-1;j>i;j--)//从个位开始找，找到最小的如果小于第i位，即交换
			{
				if(str[j]<min_c)
				{
					min_c = str[j];
					flag = j;
				}
			}
			if(str[j]!=min_c)
			{
				t = str[flag];
				str[flag] = str[i];
				str[i] = t;
				n--;
			}
		}
		printf("%s\n",str);
	}

	return 0;
}