题干：
Polycarp loves lowercase letters and dislikes uppercase ones. Once he got a string sconsisting 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.
Examples
Input
11 aaaaBaabAbA
Output
2
Input
12 zACaAbbaazzC
Output
3
Input
3 ABC
Output
0
Note
In the first example the desired positions might be 6 and 8 or 7 and 8. Positions 6and 7 contain letters 'a', position 8 contains letter 'b'. The pair of positions 1and 8 is not suitable because there is an uppercase letter 'B' between these position.
In the second example desired positions can be 7, 8 and 11. There are other ways to choose pretty set consisting of three elements.
In the third example the given string s does not contain any lowercase letters, so the answer is 0.
题目大意：
就是说给一个字符串问你有多少 在不被大写字母分割的子串中（也就是不含大写字母的子串中），小写字母的种类数最多 有多少个。
解题报告：
反正数据量很小，，随便写就行。
AC代码：
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
#include<map>
#include<vector>
#include<set>
#include<string>
#include<cmath>
#include<cstring>
#define ll long long
#define pb push_back
#define pm make_pair
#define fi first
#define se second
using namespace std;
const int MAX = 2e5 + 5;
int bk[MAX];
int cnt[505];
int n,tmp;
char s[505];
int main()
{
cin>>n;
cin>>s+1;
int len = strlen(s+1),ans = 0,tmp=0;
for(int i = 1; i<=len; i++) {
if(isupper(s[i])) {
memset(cnt,0,sizeof cnt);
ans = max(ans,tmp);
tmp=0;
continue;
}
if(cnt[s[i]] == 0) {
tmp++;
cnt[s[i]]++;
}
}
printf("%d\n",max(tmp,ans));
return 0 ;
} 总结：
注意最后输出的时候也需要取max，因为万一没进那个if(cnt[s[i]] == 0)这个if的话，，比如输入：x 那就凉凉啊。所以小地方一定要注意，，，还有一种处理办法就是每个tmp++之后就ans更新一下。。。还是那句话 一定要小范围测试一下！！
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