Aramic script（CF-#478-A）去重

A. Aramic script

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

In Aramic language words can only represent objects.

Words in Aramic have special properties:

• A word is a root if it does not contain the same letter more than once.
• A root and all its permutations represent the same object.
• The root $x$ of a word $y$ is the word that contains all letters that appear in $y$ in a way that each letter appears once. For example, the rootof "aaaa", "aa", "aaa" is "a", the root of "aabb", "bab", "baabb", "ab" is "ab".
• Any word in Aramic represents the same object as its root.

You have an ancient script in Aramic. What is the number of different objects mentioned in the script?

Input

The first line contains one integer $n$ ($1\le n\le {10}^3$) — the number of words in the script.

The second line contains $n$ words $s_1,s_2,\dots ,s_n$ — the script itself. The length of each string does not exceed ${10}^3$.

It is guaranteed that all characters of the strings are small latin letters.

Output

Output one integer — the number of different objects mentioned in the given ancient Aramic script.

Examples

input

Copy

5
a aa aaa ab abb

output

Copy

2

input

Copy

3
amer arem mrea

output

Copy

1

Note

In the first test, there are two objects mentioned. The roots that represent them are "a","ab".

In the second test, there is only one object, its root is "amer", the other strings are just permutations of "amer".

虽说是水题，但是写题解真的不清楚自己和别人的想法真不同。

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一开始我利用的是Map去重。

后来得知君玉学长直接去重。把两个写法都写一下吧。

我的做法：优点是直接找出来 按a-z的顺序

#include<bits/stdc++.h>
using namespace std;
int main()
{
    int n,ans=0;
    scanf("%d",&n);
    string tmp;
    map<string,int>mp;
    while(n--){
         cin>>tmp;
         char a[1000]={'\0'},cnt=0;
         int vis[28]={0};
         for(int i=0;i<tmp.size();i++){
            vis[tmp[i]-'a']++;
         }
         for(int i=0;i<26;i++){
            if(vis[i]!=0)
                a[cnt++]=i+'a';
         }
         string s=a;
         //cout<<s<<endl;
         if(mp[s]==0){
            mp[s]++;
            ans++;
         }
    }
    printf("%d\n",ans);
    return 0;
}

利用unique+sort+erase 函数实现去重

代码很短

#include<bits/stdc++.h>
using namespace  std;
int main()
{
    int n,ans=0;
    scanf("%d",&n);
    string s;
    map<string,int>my;
    while(n--){
        cin>>s;
        sort(s.begin(),s.end());
        s.erase((unique(s.begin(),s.end())),s.end());
        //cout<<s<<endl;
        if(my[s]==0){
            my[s]++;
            ans++;
        }
    }
    printf("%d\n",ans);
    return 0;
}