codeforces 520 Pangram

http://codeforces.com/problemset/problem/520/A

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


分析：

就是判断每一个字母是否都出现一次，大小写不计。


AC代码：

#include <stdio.h>
#include <algorithm>
#include <iostream>
#include <string.h>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <list>
#include <iomanip>
#include <vector>
#pragma comment(linker, "/STACK:1024000000,1024000000")
#pragma warning(disable:4786)

using namespace std;

const int INF = 0x3f3f3f3f;
const int MAX = 200 + 10;
const double eps = 1e-8;
const double PI = acos(-1.0);

char str;
map<char , int>ma;

int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        ma.clear();
        int i;
        for(i = 0;i <= n;i ++)
        {
            scanf("%c", &str);
            if(str >= 'A' && str <= 'Z')
                ma[str - 'A'] ++;
            else
                ma[str - 'a'] ++;
        }
        for(i = 0;i < 26;i ++)
            if(ma[i] == 0)
            {
                cout << "NO" << endl;
                break;
            }
        if(i == 26)
            cout << "YES" << endl;
    }
    return 0;
}