Description

数组中可能包含重复的数字， 求由这些数字组成的不重复字符串， 且字符串中不包含逆序对。

Input

有若干组测试数据，（1～20之间） 每一组测试数据第一行输入一个整数 n (0 ≤ n ≤ 20)， 代表 n 个数字，接下来的一行输入 n 个整数，分别为 x1， x2 … xn (1 ≤ xi ≤ 9)

Output

输出每组数字组成的不重复字符串，字符串通过逗号隔开

Sample

#0

Input

Copy

4 
3 1 2 3
8
3 4 5 1 6 1 5 5

Output

Copy

,1,12,123,1233,13,133,2,23,233,3,33
,1,11,113,1134,11345,113455,1134555,11345556,1134556,113456,11346,1135,11355,113555,1135556,113556,11356,1136,114,1145,11455,114555,1145556,114556,11456,1146,115,1155,11555,115556,11556,1156,116,13,134,1345,13455,134555,1345556,134556,13456,1346,135,1355,13555,135556,13556,1356,136,14,145,1455,14555,145556,14556,1456,146,15,155,1555,15556,1556,156,16,3,34,345,3455,34555,345556,34556,3456,346,35,355,3555,35556,3556,356,36,4,45,455,4555,45556,4556,456,46,5,55,555,5556,556,56,6

Hint

不选择任何数字时，组成空字符串

回溯求组合数然后排序

#include <iostream>
#include <vector>
#include <algorithm>
#include <set>
#include <string>
#include <cstring>
using namespace std;
void DFS(set<string> &ans, vector<int> a, int index, string path)
{
    //cout << "path=" << path << "OK" << endl;
    if (index == a.size())
    {
        ans.insert(path);
        return;
    }
    //cout << path << " ";
    
    DFS(ans, a, index+1, path);
    DFS(ans, a, index + 1, path + char(a[index] + '0'));
   
}
int main()
{
    int n;
    while (cin >> n)
    {
        vector<int> a;
        int sd;
        for (int i = 0; i < n; i++)
        {
            cin >> sd;
            a.push_back(sd);
        }
        //a.push_back(0);
        sort(a.begin(), a.end());
        set<string> s;
        vector < string>s2;
        string str="";
        DFS(s, a, 0, str);
        for (set<string>::iterator it2 = s.begin(); it2 != s.end(); it2++) {
            s2.push_back(*it2);
        }
        sort(s2.begin(), s2.end());
        for (vector<string>::iterator it = s2.begin(); it != s2.end(); it++)
        {
            if (it == s2.begin()) {
                cout << *it;
            }
            else {
                cout << "," << *it;
            }
            
        }
        cout << endl;
    }

    return 0;
}