Swap Digits

最新推荐文章于 2018-07-13 12:22:39 发布

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本文探讨了一种通过有限次数的相邻数字交换来获得最大可能数值的方法。输入包括一系列测试案例，每个案例包含一个整数K及一个初始数字串。目标是在不超过K次交换的情况下得到最大的数值组合。

Description

Now we have a number, you can swap any two adjacent digits of it, but you can not swap more than K times. Then, what is the largest probable number that we can get after your swapping?

Input

There is an integer T (1 <= T <= 200) in the first line, means there are T test cases in total.

For each test case, there is an integer K (0 <= K < 10⁶) in the first line, which has the same meaning as above. And the number is in the next line. It has at most 1000 digits, and will not start with 0.

There are at most 10 test cases that satisfy the number of digits is larger than 100.

Output

For each test case, you should print the largest probable number that we can get after your swapping.

Sample Input

Sample Output

3124
4213
4321

<span style="color:#330033;"># include <cstdio>
# include <cstring>
# include <iostream>
using namespace std;

int main()
{
    int T;
    cin>>T;
    while(T--)
    {
        int k,i,j,a=0,n,cnt=0;char c[1100],x,Max='.';
        scanf("%d%s",&k,c);
        while(k>1&&a<strlen(c))    //跳出循环条件;k=1时分开讨论;
        {
            if(a+k+1>strlen(c))   n=strlen(c);
            else   n=a+k+1;
            for(i=a+1;i<n;i++)    //找出区间内最大值；
                if(c[i]>c[a]&&c[i]>Max)  { Max=c[i];   j=i; cnt=j-a; }   //cnt记住移了多少次;
            if(Max=='.')   a++;
            else    //更新字符数组;
            {
                for(i=j;i>j-cnt;i--)    c[i]=c[i-1];
                c[a]=Max;
                Max='.';
                k-=cnt;
                a++;
            }
        }
        if(k==1)
        {
            while(c[a]>=c[a+1]&&strlen(c)>(a+1))   a++;
            if(c[a]<c[a+1])
            {
                x=c[a];
                c[a]=c[a+1];
                c[a+1]=x;
            }
        }
        puts(c);
    }
    return 0;
}</span>