Phone Numbers CodeForces - 940C（字符串.模拟）

And where the are the phone numbers?

You are given a string s consisting of lowercase English letters and an integer k. Find the lexicographically smallest string t of length k, such that its set of letters is a subset of the set of letters of s and s is lexicographically smaller than t.

It's guaranteed that the answer exists.

Note that the set of letters is a set, not a multiset. For example, the set of letters of abadaba is {a, b, d}.

String p is lexicographically smaller than string q, if p is a prefix of q, is not equal to q or there exists i, such that pi < qi and for all j < i it is satisfied that pj = qj. For example, abc is lexicographically smaller than abcd , abd is lexicographically smaller than abec, afa is not lexicographically smaller than ab and a is not lexicographically smaller than a.

Input
The first line of input contains two space separated integers n and k (1 ≤ n, k ≤ 100 000) — the length of s and the required length of t.

The second line of input contains the string s consisting of n lowercase English letters.

Output
Output the string t conforming to the requirements above.

It's guaranteed that the answer exists.

Examples
Input
3 3
abc
Output
aca
Input
3 2
abc
Output
ac
Input
3 3
ayy
Output
yaa
Input
2 3
ba
Output
baa

Note
In the first example the list of strings t of length 3, such that the set of letters of t is a subset of letters of s is as follows: aaa, aab, aac, aba, abb, abc, aca, acb, .... Among them, those are lexicographically greater than abc: aca, acb, .... Out of those the lexicographically smallest is aca.

题意：根据给出的字符串寻找按照字典序排在下一个的长度为K的字符串。
分析：分k = n , k > n , k < n三种情况。

 #include <cstdio>
#include <iostream>
#include <algorithm>
#include <cmath>
#include <string>
#include <cstring>
#include <vector>
#include <deque>
#include <stack>
#include <queue>
#include <list>
#include <map>
using namespace std;
int main()
{
    int n,m,num;
    string s,l;
    map<char,int> mp;
    scanf("%d%d",&n,&m);
    cin >> s;
    l = s;
    sort(l.begin(),l.end());
    for(int i = 0;i < n;i ++)
       mp[s[i]] = 1;
    if(m > n)
    {

        cout << s;
        for(int i = n;i < m ;i++)
            cout << l[0];
        cout << endl;
        return 0;

    }
    s[m] = 0;
    for(int i = m-1;i >= 0;i--)
    {
        if(s[i] < l[n-1])
        {
            char c;
            for(int j = 0;j < 26;j++)
            {
                c = 'a' + j;
                if(mp[c] && c > s[i])
                    break;
            }
            s[i] = c;
            break;

        }
        else
        {
            s[i] = l[0];
            continue;
        }
    }
    s.resize(m);   //重整一下长度
    cout << s << endl;

    return 0;

}