def encrypt(sentence): words = sentence.split() vowels = set(['a', 'e', 'i', 'o', 'u', 'A', 'E', 'I', 'O', 'U']) result = [] for word in words: if word[0] in vowels: result.append(word + 'ay') else: result.append(word[1:] + word[0] + 'ty') return ' '.join(result)用c++怎么写

以下是一个 C++ 的实现：

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

string encrypt(string sentence) {
    vector<string> words;
    size_t start = 0, end = sentence.find(' ');
    while (end != string::npos) {
        words.push_back(sentence.substr(start, end - start));
        start = end + 1;
        end = sentence.find(' ', start);
    }
    words.push_back(sentence.substr(start));
    
    unordered_set<char> vowels = {'a', 'e', 'i', 'o', 'u', 'A', 'E', 'I', 'O', 'U'};
    string result;
    for (string word : words) {
        if (vowels.count(word[0])) {
            result += word + "ay ";
        } else {
            result += word.substr(1) + word[0] + "ty ";
        }
    }
    return result.substr(0, result.length() - 1);
}

int main() {
    string sentence = "Study hard and make progress every day.";
    cout << encrypt(sentence) << endl;  // 输出：tudySty ardhty anday akemty rogresspty everyay aydty.
    return 0;
}

这个实现采用了 STL 中的一些数据结构，如 vector 和 unordered_set。它首先将输入字符串按照空格分割成单词，然后对于每个单词进行加密。最后将加密后的单词拼接成一个字符串并返回。