A. Equality
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
You are given a string ss of length nn, which consists only of the first kk letters of the Latin alphabet. All letters in string ss are uppercase.
A subsequence of string ss is a string that can be derived from ss by deleting some of its symbols without changing the order of the remaining symbols. For example, "ADE" and "BD" are subsequences of "ABCDE", but "DEA" is not.
A subsequence of ss called good if the number of occurences of each of the first kk letters of the alphabet is the same.
Find the length of the longest good subsequence of ss.
Input
The first line of the input contains integers nn (1≤n≤1051≤n≤105) and kk (1≤k≤261≤k≤26).
The second line of the input contains the string ss of length nn. String ss only contains uppercase letters from 'A' to the kk-th letter of Latin alphabet.
Output
Print the only integer — the length of the longest good subsequence of string ss.
Examples
input
Copy
9 3 ACAABCCAB
output
Copy
6
input
Copy
9 4 ABCABCABC
output
Copy
0
Note
In the first example, "ACBCAB" ("ACAABCCAB") is one of the subsequences that has the same frequency of 'A', 'B' and 'C'. Subsequence "CAB" also has the same frequency of these letters, but doesn't have the maximum possible length.
In the second example, none of the subsequences can have 'D', hence the answer is 00.
题意：给了n个大写字符，一个k，然后从字符串中找到从A开始到第k个字母的子序列出现了做多的个数。
题解：统计每个字母出现的个数，然后从A字母和（A+K）字母，也就是第k个字母之间字母最少出现的次数*k
c++：
#include<bits/stdc++.h>
using namespace std;
map<char,int>m;
int main()
{
int n,k,ans=1<<30;
string s;
cin>>n>>k>>s;
for(int i=0; i<n; i++)
m[s[i]]++;
for(char c='A'; c<'A'+k; c++)
ans=min(m[c],ans);
cout<<ans*k<<endl;
return 0;
}python:
一：
n,k=map(int,input().split())
s=input()
ans=[0]*k
for i in s:
ans[ord(i)-ord('A')]+=1
print(min(ans)*k)
二：
n,k=map(int,input().split())
s=input()
print(k*min(s.count(chr(ord('A')+i))for i in range(k)))