if (s.find(x) != s.end()) { return true; }
时间: 2024-02-15 10:44:48 浏览: 33
This code checks if a key 'x' exists in a set 's'. The 'find' function of a set returns an iterator to the element if it is found, else it returns an iterator to the end of the set. So, if 'x' is found in the set, the condition 's.find(x) != s.end()' will evaluate to true and the function will return true. Otherwise, it will return false.
相关问题
#include <bits/stdc++.h> using namespace std; int main() { string s,s1,s2; cin>>s>>s1>>s2; bool flag1=false,flag2=false; if(s.find(s1)!=s.npos && s.find(s2,s.find(s1)+s1.size())!=s.npos) { flag1=true; } reverse(s.begin (),s.end()); if(s.find(s1)!=s.npos && s.find(s2,s.find(s1)+s1.size())!=s.npos) { flag2=true; } if(flag1 && flag2) { cout<<"both"<<endl; }else if(flag1) { cout<<"forward"<<endl; }else if(flag2) { cout<<"backward"<<endl; }else { cout<<"fantasy"<<endl; } return 0; } 是什么意思请进行备注
这段代码是一个简单的字符串匹配程序。它接受三个输入字符串 s、s1 和 s2。程序首先检查字符串 s 是否包含 s1 和 s2,如果是,则将 flag1 设置为 true。然后,将字符串 s 反转,并再次检查是否包含 s1 和 s2,如果是,则将 flag2 设置为 true。最后,根据 flag1 和 flag2 的值,输出相应的结果。
如果 flag1 和 flag2 都为 true,则输出 "both";如果只有 flag1 为 true,则输出 "forward";如果只有 flag2 为 true,则输出 "backward";如果 flag1 和 flag2 都为 false,则输出 "fantasy"。
请注意,这段代码使用了 `<bits/stdc++.h>` 头文件,该头文件包含了大部分常用的标准库头文件。在实际开发中,更推荐使用具体的标准库头文件,而不是包含整个 `<bits/stdc++.h>`。
set<string> s = { "+", "-", "*", "/", ";", ",", "(", ")", "{", "}" }; bool inSet(const string& c) { return s.find(c) != s.end(); }
这段代码的作用是定义了一个包含特定字符的set集合s,并通过函数inSet判断参数c是否在集合s中。如果在集合s中,函数会返回true,否则返回false。可以这样演示这段代码:[^1]
```c++
#include <iostream>
#include <set>
#include <string>
using namespace std;
set<string> s = { "+", "-", "*", "/", ";", ",", "(", ")", "{", "}" };
bool inSet(const string& c) {
return s.find(c) != s.end();
}
int main() {
string test1 = "+";
string test2 = "!";
if (inSet(test1)) {
cout << test1 << " is in the set." << endl; // 输出:+ is in the set.
}
if (!inSet(test2)) {
cout << test2 << " is not in the set." << endl; // 输出:! is not in the set.
}
return 0;
}
```
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import re,tkinter,requests,threading,tqdm as tt root = tkinter.Tk() root.title('在线视频解析') root.geometry('500x590+550+350') headers = { 'User-Agent': 'Mozilla/5.0 (Windows NT 10.0; Win64; x64; rv:109.0) Gecko/20100101 Firefox/115.0'} ac = tkinter.Listbox(root, width=50, height=20, font=('黑体', 12)) ac.grid(row=2, columnspan=10, sticky="n" + "s" + "w" + "e") def sousuo(): i = b1.get() ac.delete(0, 'end') def extract_music_info(content): p = '|' content = re.sub(p, '', content, flags=re.S) pattern = re.compile('subject.*?href="(.*?)">(.*?)
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