A. Generate Login（string）

A. Generate Login（string）

time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
The preferred way to generate user login in Polygon is to concatenate a prefix of the user’s first name and a prefix of their last name, in that order. Each prefix must be non-empty, and any of the prefixes can be the full name. Typically there are multiple possible logins for each person.

You are given the first and the last name of a user. Return the alphabetically earliest login they can get (regardless of other potential Polygon users).

As a reminder, a prefix of a string s is its substring which occurs at the beginning of s: “a”, “ab”, “abc” etc. are prefixes of string “{abcdef}” but “b” and ‘bc” are not. A string a is alphabetically earlier than a string b, if a is a prefix of b, or a and b coincide up to some position, and then a has a letter that is alphabetically earlier than the corresponding letter in b: “a” and “ab” are alphabetically earlier than “ac” but “b” and “ba” are alphabetically later than “ac”.

Input
The input consists of a single line containing two space-separated strings: the first and the last names. Each character of each string is a lowercase English letter. The length of each string is between 1 and 10, inclusive.

Output
Output a single string — alphabetically earliest possible login formed from these names. The output should be given in lowercase as well.

Examples
input
harry potter
output
hap
input
tom riddle
output
tomr

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
#include <set>
#include <map>
#include <vector>
#define INF 0x3f3f3f3f
using namespace std;
vector<string>ans;
int main()
{
    string a, b;
    cin>>a>>b;
    ans.clear();//清空
    int numa = a.length();//获取长度
    int numb = b.length();
    for(int i=1; i<=numa; i++)
    {
        for(int j=1; j<=numb; j++)
        {
            string temp = a.substr(0, i)+b.substr(0, j);//连接
            ans.push_back(temp);
        }
    }
    sort(ans.begin(), ans.end());//排序
    cout<<ans[0]<<endl;
    return 0;
}