题目：
You are given a string s consisting of n lowercase Latin letters. Polycarp wants to remove exactly k characters (k≤n) from the string s. Polycarp uses the following algorithm k times:
if there is at least one letter ‘a’, remove the leftmost occurrence and stop the algorithm, otherwise go to next item; if there is at least one letter ‘b’, remove the leftmost occurrence and stop the algorithm, otherwise go to next item; … remove the leftmost occurrence of the letter ‘z’ and stop the algorithm. This algorithm removes a single letter from the string. Polycarp performs this algorithm exactly k times, thus removing exactly k characters.
Help Polycarp find the resulting string.
Input The first line of input contains two integers n and k (1≤k≤n≤4⋅105) — the length of the string and the number of letters Polycarp will remove.
The second line contains the string s consisting of n lowercase Latin letters.
Output Print the string that will be obtained from s after Polycarp removes exactly k letters using the above algorithm k times.
If the resulting string is empty, print nothing. It is allowed to print nothing or an empty line (line break).
Examples Input 15 3 cccaabababaccbc Output cccbbabaccbc Input 15 9 cccaabababaccbc Output cccccc Input 1 1 u Output （注意，这里没有输出）
题目大意：
给你一个指定长度的均由小写字母组成的字符串，然后从中删除指定个数的字符，从a开始删，a删完了删b……一直删到符合要求为止。
刚开始理解错题意了，以为删z的时候只删第一个就可以了，没想到也是删到符合要求为止。注意，题目给的意思就是只要a>=1就删，即把a删完才继续删除b，而不是剩一个a不删！
AC代码：（略长，因为26个字母我都列出来了，各位大佬有更好的方法的话欢迎评论）
#include<iostream>
#include<string>
#include<algorithm>
using namespace std
;
int main()
{
int n
=0,k
=0;
cin
>>n
>>k
;
char a
[500000];
scanf("%s",&a
);
for(int i
=0;i
<n
;i
++)
{
if(a
[i
]=='a'&&k
>0)
{
a
[i
]='*';
k
--;
}
}
for(int i
=0;i
<n
;i
++)
{
if(a
[i
]=='b'&&k
>0)
{
a
[i
]='*';
k
--;
}
}
for(int i
=0;i
<n
;i
++)
{
if(a
[i
]=='c'&&k
>0)
{
a
[i
]='*';
k
--;
}
}
for(int i
=0;i
<n
;i
++)
{
if(a
[i
]=='d'&&k
>0)
{
a
[i
]='*';
k
--;
}
}
for(int i
=0;i
<n
;i
++)
{
if(a
[i
]=='e'&&k
>0)
{
a
[i
]='*';
k
--;
}
}
for(int i
=0;i
<n
;i
++)
{
if(a
[i
]=='f'&&k
>0)
{
a
[i
]='*';
k
--;
}
}
for(int i
=0;i
<n
;i
++)
{
if(a
[i
]=='g'&&k
>0)
{
a
[i
]='*';
k
--;
}
}
for(int i
=0;i
<n
;i
++)
{
if(a
[i
]=='h'&&k
>0)
{
a
[i
]='*';
k
--;
}
}
for(int i
=0;i
<n
;i
++)
{
if(a
[i
]=='i'&&k
>0)
{
a
[i
]='*';
k
--;
}
}
for(int i
=0;i
<n
;i
++)
{
if(a
[i
]=='j'&&k
>0)
{
a
[i
]='*';
k
--;
}
}
for(int i
=0;i
<n
;i
++)
{
if(a
[i
]=='k'&&k
>0)
{
a
[i
]='*';
k
--;
}
}
for(int i
=0;i
<n
;i
++)
{
if(a
[i
]=='l'&&k
>0)
{
a
[i
]='*';
k
--;
}
}
for(int i
=0;i
<n
;i
++)
{
if(a
[i
]=='m'&&k
>0)
{
a
[i
]='*';
k
--;
}
}
for(int i
=0;i
<n
;i
++)
{
if(a
[i
]=='n'&&k
>0)
{
a
[i
]='*';
k
--;
}
}
for(int i
=0;i
<n
;i
++)
{
if(a
[i
]=='o'&&k
>0)
{
a
[i
]='*';
k
--;
}
}
for(int i
=0;i
<n
;i
++)
{
if(a
[i
]=='p'&&k
>0)
{
a
[i
]='*';
k
--;
}
}
for(int i
=0;i
<n
;i
++)
{
if(a
[i
]=='q'&&k
>0)
{
a
[i
]='*';
k
--;
}
}
for(int i
=0;i
<n
;i
++)
{
if(a
[i
]=='r'&&k
>0)
{
a
[i
]='*';
k
--;
}
}
for(int i
=0;i
<n
;i
++)
{
if(a
[i
]=='s'&&k
>0)
{
a
[i
]='*';
k
--;
}
}
for(int i
=0;i
<n
;i
++)
{
if(a
[i
]=='t'&&k
>0)
{
a
[i
]='*';
k
--;
}
}
for(int i
=0;i
<n
;i
++)
{
if(a
[i
]=='u'&&k
>0)
{
a
[i
]='*';
k
--;
}
}
for(int i
=0;i
<n
;i
++)
{
if(a
[i
]=='v'&&k
>0)
{
a
[i
]='*';
k
--;
}
}
for(int i
=0;i
<n
;i
++)
{
if(a
[i
]=='w'&&k
>0)
{
a
[i
]='*';
k
--;
}
}
for(int i
=0;i
<n
;i
++)
{
if(a
[i
]=='x'&&k
>0)
{
a
[i
]='*';
k
--;
}
}
for(int i
=0;i
<n
;i
++)
{
if(a
[i
]=='y'&&k
>0)
{
a
[i
]='*';
k
--;
}
}
for(int i
=0;i
<n
;i
++)
{
if(a
[i
]=='z'&&k
>0)
{
a
[i
]='*';
k
--;
}
}
for(int i
=0;i
<n
;i
++)
{
if(a
[i
]!='*')
printf("%c",a
[i
]);
}
return 0;
}
