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?
标签：des style io ar color os sp for strong
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
3
2
1234
4
1234
1
4321Sample 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>
标签：des style io ar color os sp for strong
原文地址：http://blog.csdn.net/rechard_chen/article/details/41458347
