codeforces——443A——Anton and Letters

最新推荐文章于 2021-06-08 08:11:21 发布

lose__way

最新推荐文章于 2021-06-08 08:11:21 发布

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分类专栏： codeforces 文章标签： codeforces ACM c语言

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本文介绍了一个简单的编程问题解决方案，即从一组可能重复的小写英文字符中统计不同的字母数量。通过使用C++语言实现，该算法有效地解决了问题，并提供了一种快速的方法来计算字符串中不重复的字母总数。

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Recently, Anton has found a set. The set consists of small English letters. Anton carefully wrote out all the letters from the set in one line, separated by a comma. He also added an opening curved bracket at the beginning of the line and a closing curved bracket at the end of the line.

Unfortunately, from time to time Anton would forget writing some letter and write it again. He asks you to count the total number of distinct letters in his set.

Input

The first and the single line contains the set of letters. The length of the line doesn't exceed 1000. It is guaranteed that the line starts from an opening curved bracket and ends with a closing curved bracket. Between them, small English letters are listed, separated by a comma. Each comma is followed by a space.

Output

Print a single number — the number of distinct letters in Anton's set.

Examples

Input

{a, b, c}

Output

Input

{b, a, b, a}

Output

Input

{}

Output

数不同字符串中不同字母的数量

水题

#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<cmath>
#include<cstring>
using namespace std;
int func1(int a,int b);
int main()
{
    char c;
    while(~scanf("%c",&c))
    {
        int a[26];
        memset(a,0,sizeof(a));
        while(1)
        {
            c=getchar();
            if(c=='}')
            {
                getchar();
                break;
            }
            a[c-97]=1;
            c=getchar();
            getchar();
            if(c=='}')
                break;
        }
        int ans=0;
        for(int i=0; i<26; i++)
            if(a[i]==1)
                ans++;
        cout<<ans<<endl;
    }
    return 0;
}