题目链接：http://acm.fzu.edu.cn/problem.php?pid=2111
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!
InputThe 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 Sample Input Sample Output Source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;
}
FZU Problem 2111 Min Number （数学啊 ）
原文：http://blog.csdn.net/u012860063/article/details/41752751