std::vector<int> repeatNumbers(sentence.size() + 1, 0);
时间: 2024-04-13 21:27:31 浏览: 18
`std::vector<int> repeatNumbers(sentence.size() + 1, 0);` 是一个变量声明和初始化的语句。其中,`std::vector<int>` 是 C++ 标准库中的一个容器类,用于存储一组具有相同类型的元素,而 `repeatNumbers` 是该容器类的一个对象。
这条语句创建了一个名为 `repeatNumbers` 的 `std::vector<int>` 类型的对象,并将其初始化为指定大小为 `sentence.size() + 1` 的向量,并且所有元素的初始值都设置为 0。
换句话说,`repeatNumbers` 是一个整数向量,其大小为 `sentence.size() + 1`,并且初始时所有元素的值都为 0。这个向量可以用来存储一组整数,并进行相关操作,比如访问、修改或者遍历等。
相关问题
用C++重写def calc(sentence,DAG,idx,route): N = len(sentence) route[N] = (0.0,'') for idx in xrange(N-1,-1,-1): candidates = [ ( FREQ.get(sentence[idx:x+1],min_freq) + route[x+1][0],x ) for x in DAG[idx] ] route[idx] = max(candidates)
Sure! Here's the C++ code equivalent of the given Python code:
```cpp
#include <iostream>
#include <vector>
#include <map>
#include <algorithm>
using namespace std;
double calc(const string& sentence, const vector<vector<int>>& DAG, int idx, map<int, pair<double, string>>& route) {
int N = sentence.length();
route[N] = make_pair(0.0, "");
for (int idx = N-1; idx >= 0; idx--) {
vector<pair<double, int>> candidates;
for (int x : DAG[idx]) {
double freq = FREQ[sentence.substr(idx, x-idx+1)];
candidates.push_back(make_pair(freq + route[x+1].first, x));
}
auto max_candidate = max_element(candidates.begin(), candidates.end());
route[idx] = make_pair(max_candidate->first, sentence.substr(idx, max_candidate->second - idx + 1));
}
return route[0].first;
}
int main() {
string sentence = "example sentence";
vector<vector<int>> DAG = { {1, 2}, {3}, {4}, {5}, {}, {6} };
int idx = 0;
map<int, pair<double, string>> route;
double result = calc(sentence, DAG, idx, route);
cout << "Result: " << result << endl;
return 0;
}
```
Please note that you need to define the `FREQ` map and initialize it with the appropriate values before using it in the `calc` function.
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:
```cpp
#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。
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