CodeForces 443A Anton and Letters

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链接：http://codeforces.com/problemset/problem/443/A

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

Input

{b, a, b, a}

Output

Input

{}

Output

大意——给你一个集合，集合里面只含有小写英文字母。我们是用字符串表示这个集合的，表示时集合以前半个大括号开始并且以后半个大括号结束，大括号里面是小写英文字母，以英文逗号分隔。现在要你计算出该集合里有几个互不同的小写英文字母。

思路——字符串最长也就1000，一个个枚举即可。用一个数组记录一个字母有无，当发现这个串里面有小写英文字母时，就将表示这个英文字母的数组元素置为有。最后将该数组枚举一遍，并且记录字母个数即得到答案。

复杂度分析——时间复杂度:O(len),空间复杂度:O(len)

附上AC代码：

#include <iostream>
#include <cstdio>
#include <string>
#include <cmath>
#include <iomanip>
#include <ctime>
#include <climits>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <queue>
#include <vector>
#include <set>
#include <map>
//#pragma comment(linker, "/STACK:102400000, 102400000")
using namespace std;
typedef unsigned int li;
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
const double pi = acos(-1.0);
const double e = exp(1.0);
const double eps = 1e-8;
const short maxlen = 1005;
char str[maxlen];int main()
{ios::sync_with_stdio(false);while (gets(str)){bool flag[30]={0};short len = strlen(str);for (short i=0; i<len; i++)if (islower(str[i]))flag[str[i]-'a'] = 1;short sum = 0;for (short i=0; i<26; i++)if (flag[i])sum++;printf("%hd\n", sum);}return 0;
}

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