Codeforces Round #455 Div.2 909A

最新推荐文章于 2021-01-12 23:25:09 发布

原创最新推荐文章于 2021-01-12 23:25:09 发布 · 302 阅读

0 ·

CC 4.0 BY-SA版权

小朋友学Codeforces 专栏收录该内容

24 篇文章

订阅专栏

本文解析了CodeForces竞赛中一道A级题目，通过具体示例详细阐述了解题思路和算法实现过程。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Problem

http://codeforces.com/contest/909/problem/A

Analysis

According to the problem statements, you need to consider all the characters in the first name, and only the first character in the last name.

(1) Take “harry potter”for example
① You got“hp”
② You got “hap”, which is less than “hp”
③ You got “harp”,which is greater than “hap”
So, the answer is “hap”

(2) Take“ertuyivhfg v”for example
① You got “ev”
② You got “erv”, which is less than “ev”
③ You got “ertv”, which is less than “erv”
④ You got “ertuv”, which is less than “ertv”
⑤ You got “ertuyv”, which is greater than “ertuv”
So, the answer is “ertuv”

Code

#include<iostream>
#include<string>
using namespace std;

int main() 
{
    char first[10], last[10];
    cin >> first;
    cin >> last;

    string pre1, pre2, temp, result;
    pre1 = "";
    pre1 += first[0];
    pre2 = "";
    pre2 += last[0];
    result = pre1 + pre2;

    for (int i = 1; first[i]; i++) 
    {
        pre1 += first[i];
        temp = pre1 + pre2;

        if (temp < result)
        {
            result = temp;
        }
    }

    cout << result;
}