CodeForces 544A Set of Strings

http://codeforces.com/problemset/problem/544/A

Set of Strings

You are given a string q. A sequence of k strings s1, s2, ..., sk is called beautiful, if the concatenation of these strings is string q(formally, s1 + s2 + ... + sk = q) and the first characters of these strings are distinct.

Find any beautiful sequence of strings or determine that the beautiful sequence doesn't exist.

Input

The first line contains a positive integer k (1 ≤ k ≤ 26) — the number of strings that should be in a beautiful sequence.

The second line contains string q, consisting of lowercase Latin letters. The length of the string is within range from 1 to 100, inclusive.

Output

If such sequence doesn't exist, then print in a single line "NO" (without the quotes). Otherwise, print in the first line "YES" (without the quotes) and in the next k lines print the beautiful sequence of strings s1, s2, ..., sk.

If there are multiple possible answers, print any of them.

Sample test(s)

input

1
abca

output

YES
abca

input

2
aaacas

output

YES
aaa
cas

input

4
abc

output

NO

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <cstdlib>
#include <limits>
#include <queue>
#include <stack>
#include <vector>
#include <map>

using namespace std;
typedef long long LL;

#define N 110
#define INF 0x3f3f3f3f
#define PI acos (-1.0)
#define EPS 1e-8
#define met(a, b) memset (a, b, sizeof (a))

int main ()
{
    int n, a[N];
    char str[N];

    while (scanf("%d %s", &n, str) != EOF)
    {
        met (a, 0);

        for (int i=0; str[i]; i++)
            a[str[i]-'a']++;

        int cnt = 0, vis[N];

        for (int i=0; i<26; i++)
            if (a[i]) cnt++;
        met (vis, 0);//标记一个字符是否在前边的字符串中的第一个字符中出现过

        if (cnt >= n)
        {
            puts ("YES");

            int k = 0, flag = 1;
            printf ("%c", str[0]);
            vis[str[k]-'a'] = 1;

            for (int i=1; str[i]; i++)
            {
                if (flag == n)//当是最后一个字符的时候全部输出
                {
                    for (int j=i; str[j]; j++)
                        printf ("%c", str[j]);

                    break;
                }
                if ((a[str[i]-'a'] != a[str[k]-'a'] || str[i] != str[k]) && !vis[str[i]-'a'])
                {
                    k = i;//记录该字符串的第一个字符
                    vis[str[i]-'a'] = 1;
                    flag++;
                    puts ("");
                }

                printf ("%c", str[i]);
            }
            puts ("");
        }
        else puts ("NO");
    }
    return 0;
}