FZU Problem 2111 Min Number （数学啊 ）

题目链接：http://acm.fzu.edu.cn/problem.php?pid=2111

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

Source

“高教社杯”第三届福建省大学生程序设计竞赛

题意：

给出一个数字，求经过n次交换后最小的数字！

代码如下：

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
int main()
{
    int t;
    int n;
    char s[117];
    scanf("%d",&t);
    while(t--)
    {
        scanf("%s%d",s,&n);
        int len = strlen(s);
        for(int i = 0; i < n; i++)
        {
            int tt = s[i]-'0';
            int pos = 0, minn = tt;
            for(int j = i+1; j < len; j++)
            {
                if(i == 0 && s[j] == '0')//首位
                    continue;
                if(s[j]-'0' < minn)
                {
                    minn = s[j]-'0';
                    pos = j;
                }
            }
            if(pos != 0)
                swap(s[i],s[pos]);
            else//无效操作
                n++;
            if(n >= len)
                n = len-1;
        }
        printf("%s\n",s);
    }
    return 0;
}