[Codeforces 864B]Polycarp and Letters

Description

Polycarp loves lowercase letters and dislikes uppercase ones. Once he got a string s consisting only of lowercase and uppercase Latin letters.

Let A be a set of positions in the string. Let's call it pretty if following conditions are met:

letters on positions from A in the string are all distinct and lowercase;
there are no uppercase letters in the string which are situated between positions from A (i.e. there is no such j that s[j] is an uppercase letter, and a1 < j < a2 for some a1 and a2 from A).

Write a program that will determine the maximum number of elements in a pretty set of positions.

Input

The first line contains a single integer n (1 ≤ n ≤ 200) — length of string s.

The second line contains a string s consisting of lowercase and uppercase Latin letters.

Output

Print maximum number of elements in pretty set of positions for string s.

Sample Input

11aaaaBaabAbA

Sample Output

HINT

In the first example the desired positions might be 6 and 8 or 7 and 8. Positions 6 and 7 contain letters 'a', position 8 contains letter 'b'. The pair of positions 1 and 8 is not suitable because there is an uppercase letter 'B' between these position.

题解

在一个字符串中找一段，使其中没有大写字母，并且小写字母种数最多。

考虑$n<=200$，直接暴力枚举。

I think there is no need to tell you how to solve this problem...

 //It is made by Awson on 2017.9.29
 #include <set>
 #include <map>
 #include <cmath>
 #include <ctime>
 #include <queue>
 #include <stack>
 #include <vector>
 #include <cstdio>
 #include <string>
 #include <cstring>
 #include <cstdlib>
 #include <iostream>
 #include <algorithm>
 #define LL long long
 #define Max(a, b) ((a) > (b) ? (a) : (b))
 #define Min(a, b) ((a) < (b) ? (a) : (b))
 #define sqr(x) ((x)*(x))
 #define lowbit(x) ((x)&(-(x)))
 using namespace std;
 void read(int &x) {
     ;
     ); ch = getchar());
     ; isdigit(ch); x = (x<<)+(x<<)+ch-, ch = getchar());
     x *= -*flag;
 }

 int n, ans;
 ];
 ];

 void work() {
     read(n);
     ; i <= n; i++)
         cin>>ch[i];
     ; i <= n; i++)
         for (int j = i; j <= n; j++) {
             memset(judge, , sizeof(judge));
             ;
             for (int k = i; k <= j; k++)
                 if (ch[k] >= 'A' && ch[k] <= 'Z') {
                     flag = ;
                     break;
                 }
                 ;
             if (flag) continue;
             ;
             ; k < ; k++)
                 cnt += judge[k];
             ans = Max(ans, cnt);
         }
     printf("%d\n", ans);
 }
 int main() {
     work();
     ;
 }

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