week4-[字符数组]月份

week4-[字符数组]月份

题目描述

1月到12月的英语单词分别为：January、February、March、April、May、June、July、August、September、October、November、December。

现在给定其中11个月份的英语单词，请求出没有出现的那个月份对应的英语单词。

输入格式

输入11行，每行读入一个字符串，为某个月份对应的英语单词。保证输入的11个字符串互不相同。

输出格式

输出1行，该行包括一个字符串，为在输入中没有出现的那个月份对应的英语单词。

样例 #1

样例输入 #1

August
May
June
July
September
January
February
March
November
December
April

样例输出 #1

October

样例 #2

样例输入 #2

January
October
July
May
December
August
September
June
April
February
March

样例输出 #2

November

样例 #3

样例输入 #3

August
April
February
May
January
October
July
November
December
March
September

样例输出 #3

June

这题比较经典，本质就是 找出全集（12个月份）中缺失的那个元素。

---

解题思路

1. 我们先准备一个完整的月份数组：

   string months[12] = {"January","February","March","April","May","June","July","August","September","October","November","December"
   };
   

2. 用一个 哈希表（map/set） 来存储输入的 11 个字符串。

3. 遍历 months[12]，查找哪个不在输入集合里，就是答案。

---

C++参考代码

#include <bits/stdc++.h>
using namespace std;int main() {ios::sync_with_stdio(false);cin.tie(nullptr);// 1. 完整月份表string months[12] = {"January","February","March","April","May","June","July","August","September","October","November","December"};// 2. 输入集合set<string> s;for (int i = 0; i < 11; i++) {string str;cin >> str;s.insert(str);}// 3. 找缺失的月份for (int i = 0; i < 12; i++) {if (s.find(months[i]) == s.end()) {cout << months[i] << "\n";break;}}return 0;
}

---

✅ 时间复杂度：O(12 + 11 log 11)，常数极小，可以忽略。
✅ 空间复杂度：O(11)。