codeforces 520 A. Pangram (暴力)

A. Pangram

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

A word or a sentence in some language is called a pangram if all the characters of the alphabet of this language appear in it at least once. Pangrams are often used to demonstrate fonts in printing or test the output devices.

You are given a string consisting of lowercase and uppercase Latin letters. Check whether this string is a pangram. We say that the string contains a letter of the Latin alphabet if this letter occurs in the string in uppercase or lowercase.

Input

The first line contains a single integer n (1 ≤ n ≤ 100) — the number of characters in the string.

The second line contains the string. The string consists only of uppercase and lowercase Latin letters.

Output

Output "YES", if the string is a pangram and "NO" otherwise.

Sample test(s)

input

12
toosmallword

output

NO

input

35
TheQuickBrownFoxJumpsOverTheLazyDog

output

YES

给一个字符串，问这个字符串中是否26个字母都出现过（大小写只出现一个就算出现过）
开个布尔数组，扫一遍即可。
嘛，做两道水题放松下==
反正也是要清的。

/*************************************************************************
    > File Name: code/cf/#295/A.cpp
    > Author: 111qqz
    > Email: rkz2013@126.com 
    > Created Time: 2015年08月17日 星期一 04时05分12秒
 ************************************************************************/

#include<iostream>
#include<iomanip>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<string>
#include<map>
#include<set>
#include<queue>
#include<vector>
#include<stack>
#define y0 abc111qqz
#define y1 hust111qqz
#define yn hez111qqz
#define j1 cute111qqz
#define tm crazy111qqz
#define lr dying111qqz
using namespace std;
#define REP(i, n) for (int i=0;i<int(n);++i)  
typedef long long LL;
typedef unsigned long long ULL;
const int inf = 0x7fffffff;
bool v[30];
int main()
{
    int n;
    scanf("%d",&n);
    memset(v,false,sizeof(v));
    string st;
    cin>>st;
    for ( int i = 0 ; i < n ; i ++)
    {
    if (islower(st[i]))
    {
        v[st[i]-'a'] = true;
    }
    else
    {
        v[st[i]-'A'] = true;
    }
    }
    for ( int i = 0 ; i < 26 ; i++)
    {
    if (!v[i])
    {
        
        cout<<"NO"<<endl;
        return 0;
    }
    }
    cout<<"YES"<<endl;
  
    return 0;
}