time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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.

Sample test(s)
input
{a, b, c}
output
3
input
{b, a, b, a}
output
2
input
{}
output
0

解题说明:此题事实上就是考察C++中的set,把输入处理好就可以。

#include<iostream>
#include<cstdio>
#include<cmath>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include <set>
using namespace std; int main()
{
string s;
set<char> t;
while (cin>>s)
{
if (s[0]!='{')
{
t.insert(s[0]);
}
else if (s[1]!='}')
{
t.insert(s[1]);
}
}
cout<<t.size()<<endl;
return 0;
}

A. Anton and Letters的更多相关文章

  1. cf443A Anton and Letters

    A. Anton and Letters time limit per test 2 seconds memory limit per test 256 megabytes input standar ...

  2. Anton and Letters

    Anton and Letters time limit per test 2 seconds memory limit per test 256 megabytes input standard i ...

  3. Y - Anton and Letters

    Problem description Recently, Anton has found a set. The set consists of small English letters. Anto ...

  4. Codeforces Round #253 (Div. 2) A. Anton and Letters

    题目很简单,只需要注意带空格的输入用getline即可 #include <iostream> #include <vector> #include <algorithm ...

  5. Codeforces Round 253 (Div. 2)

    layout: post title: Codeforces Round 253 (Div. 2) author: "luowentaoaa" catalog: true tags ...

  6. Codeforces Round #379 (Div. 2) A. Anton and Danik 水题

    A. Anton and Danik 题目连接: http://codeforces.com/contest/734/problem/A Description Anton likes to play ...

  7. C - Anton and Danik

    Problem description Anton likes to play chess, and so does his friend Danik. Once they have played n ...

  8. 【77.39%】【codeforces 734A】Anton and Danik

    time limit per test1 second memory limit per test256 megabytes inputstandard input outputstandard ou ...

  9. [LeetCode] Remove Duplicate Letters 移除重复字母

    Given a string which contains only lowercase letters, remove duplicate letters so that every letter ...

随机推荐

  1. iphone6闪存检测

    iPhone6自从发布以后一直又不少的诟病和非议,比如一机难求,容易掰弯,程序崩溃等, 甚至传出了苹果将要召回这些问题设备,最近有人终于查出了iPhone6安装大量程序后崩溃的原因,原因就是大容量的i ...

  2. execute immediate的简单用法(oracle)

    直接上示例代码: create or replace procedure proc_test( --参数区域 ) is --变量区域 --sql脚本 v_sql ) :=''; --记录学生数量 v_ ...

  3. N种方法妙讲LIS算法

    LIS算法经典汇总 假设存在一个序列d[1..9] = 2 1 5 3 6 4 8 9 7,可以看出来它的LIS长度为5.下面一步一步试着找出它.我们定义一个序列B,然后令 i = 1 to 9 逐个 ...

  4. MFC学习之程序执行过程梳理

    *首先利用全局变量对象theApp启动应用程序.这是由于这个全局对象,基类CWinApp中this的指针才干指向这个对象.假设没有这个全局对象,程序在编译时不会出错,但在执行时就会出错. *调用全局应 ...

  5. Gridview 多重表头 (二)

    多重表头之排序 这是个有点忧桑的故事...Cynthia告诉我,研究一个问题,我们不可能有超过一天的时间... 结果好好几天过去鸟~~还没有完成... 由于不再使用Gridview自带的表头行,于是无 ...

  6. SqlServer判断数据库、表、存储过程、函数是否存在

    假设场景是: 需要给一个脚本给客户更新, 这个对象可能存在或不存在 -- 更新存储过程 USE [数据库名] GO SET ANSI_NULLS ON GO SET QUOTED_IDENTIFIER ...

  7. android——ObjectAnimator动画

    在新的android sdk中谷歌为我们提供了新的动画实现方式.化繁为简.将以前的animation动画进一步封装,使用起来更加方便. 先来看XML文件: <RelativeLayout xml ...

  8. 读取webconfig里面的appSetting和connectionString

    <appSettings> <add key="SiteURL" value="http://moss2007:7000" /> < ...

  9. for循环删除集合陷阱

    首先看下面的代码: import java.util.LinkedList;import java.util.List; public class DeleteCollection {         ...

  10. Java 面向对象(转)

    转自:http://blog.sina.com.cn/s/blog_83c5190f01010ate.html 1.我们可以把JAVA中的类分为以下三种: 类:使用class定义且不含有抽象方法的类. ...