Pasha has a positive integer a without leading zeroes. Today he decided that the number is too small and he should make it larger. Unfortunately, the only operation Pasha can do is to swap two adjacent decimal digits of the integer.
Help Pasha count the maximum number he can get if he has the time to make at most k swaps.
The single line contains two integers a and k (1 ≤ a ≤ 1018; 0 ≤ k ≤ 100).
Print the maximum number that Pasha can get if he makes at most k swaps.
1990 1
9190
300 0
300
1034 2
3104
9090000078001234 6
9907000008001234
Pasha has a positive integer a without leading zeroes. Today he decided that the number is too small and he should make it larger. Unfortunately, the only operation Pasha can do is to swap two adjacent decimal digits of the integer.
Help Pasha count the maximum number he can get if he has the time to make at most k swaps.
The single line contains two integers a and k (1 ≤ a ≤ 1018; 0 ≤ k ≤ 100).
Print the maximum number that Pasha can get if he makes at most k swaps.
1990 1
9190
300 0
300
1034 2
3104
9090000078001234 6
9907000008001234
其实 很简单,尽量交换前面的数字可以保证最优嘛
#include<iostream> #include<cstdio> #include<cstring> #include<string> #include<cmath> #include<algorithm> #include<cstdlib> using namespace std; string s; int k; int main() { cin>>s>>k; for(int i=0;i<s.size();i++) { int b=i; for(int j=i+1;j<s.size()&&j-i<=k;j++) { if(s[j]>s[b]) b=j; } k-=b-i; for(int j=b;j>i;j--) swap(s[j],s[j-1]); } cout<<s<<endl; return 0; }