B. Pasha Maximizes
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
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.
Input
The single line contains two integers a and
k (1 ≤ a ≤ 1018; 0 ≤ k ≤ 100).
Output
Print the maximum number that Pasha can get if he makes at most
k swaps.
Sample test(s)
Input
1990 1
Output
9190
Input
300 0
Output
300
Input
1034 2
Output
3104
Input
9090000078001234 6
Output
9907000008001234
简单贪心应用
#include<map>
#include<set>
#include<list>
#include<queue>
#include<stack>
#include<vector>
#include<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
int cmp(int a, int b)
{
return a > b;
}
int arr[20];
int b[20];
int main()
{
__int64 a;
int k;
while(~scanf("%I64d", &a))
{
scanf("%d", &k);
__int64 ca = a;
int dig = 0;
while(ca)
{
ca /= 10;
dig++;
}
for(int i = dig - 1; i >= 0; i--)
{
arr[i] = a % 10;
b[i] = arr[i];
a /= 10;
}
sort(b, b + dig, cmp);
int s = 0;
while(k > 0)
{
int maxs = -1, pos = 0;
int len = min(s + k, dig - 1);
for(int i = s; i <= len; i++)
{
if(maxs < arr[i])
{
maxs = arr[i];
pos = i;
}
}
for(int i = pos; i > s; i--)
{
arr[i] ^= arr[i - 1];
arr[i - 1] ^= arr[i];
arr[i] ^= arr[i - 1];
}
k -= pos - s;
s++;
}
for(int i = 0; i < dig; i++)
printf("%d", arr[i]);
printf("\n");
}
return 0;
}