Min Number
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
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
char a[10000000],b[10000000];
int main()
{
int t,l,i,j;
scanf("%d",&t);
while(t--)
{
int m;
scanf("%s%d",a,&m);
l=strlen(a);
if(m==0)
{
printf("%s\n",a);
}
else
{
for(i=0; i<l; i++)
{
b[i]=a[i];
}
sort(b,b+l);//排序
int h=0;
char we=a[0];
for(i=0; i<l; i++)//把第一个数的大小先排好
{
if(b[i]!='0')
{
if(a[0]>b[i]&&b[i]!=-1)
{
for(j=0; j<l; j++)
{
if(a[j]==b[i])//找到原来的b[i]的位置
{
b[i]=-1;
int f;
f=j;
int s;
s=a[0];//交换
a[0]=a[f];
a[f]=s;
m--;
}
}
}
else if(a[0]==b[i]&&b[i]!=-1)
{
h++;
b[i]=-1;
if(h>1)
{
//printf("&&&%d\n",h);
for(int v=1; v<h; v++)
{
b[i]=we;
//printf("%c %c %d\n",b[i],we,i);
}
}
else
{
b[i]=-1;
}
}
}
}
int p=1;//从后一个开始排
int q,w;
for(j=0; j<l; j++)
{
if(a[p]>b[j]&&b[j]!=-1)
{
for(int k=p+1; k<l; k++)//注意此句，从p的后一个开始找b[j]原来的位置
{
if(a[k]==b[j])//找到之后记住下标,然后交换位置
{
b[j]=-1;
int r;
r=k;
int q;
q=a[p];
a[p]=a[r];
a[r]=q;
m--;
p++;
if(m<=0||p>=l)
{
break;
}
}
}
}
if(a[p]<=b[j]&&b[j]!=-1)
{
b[j]=-1;
p++;
if(p>=l||m<=0)
{
break;
}
}
}
for(i=0; i<l; i++)
{
printf("%c",a[i]);
}
printf("\n");
}
}
}