L1-059 ring stupid bell

A problem that is not very difficult , But pay attention to the pit ！

Topic link

#include<iostream>
#include<string>
using namespace std;
int main() {
    
	string s;
	int N;
	cin >> N;
	getchar();
	for (int i = 0; i < N; i++) {
    
		getline(cin, s);
		int pos=s.find(",");
		string ong1,ong2;
		if (pos < 3) {
    	// Pay attention to the length of the first half 
			cout << "Skipped";
			if (i != N - 1)	cout << endl;
			continue;
		}
		ong1 = s.substr(pos - 3, 3);
		ong2 = s.substr(s.size() - 4, 3);
		if (ong1 != "ong" || ong2 != "ong")	cout << "Skipped";
		else {
    
			for (int j = 0; j < 3; j++) {
    
				int pos1 = s.rfind(" ");	// Delete from back to front 3 A word , A space represents a word 
				s.erase(pos1);
			}
			s += " qiao ben zhong.";
			cout << s;
		}
		if (i != N - 1)	cout << endl;
	}
	return 0;
}

Description of the input format of the topic ：

The input first gives a no more than 20 The positive integer N. And then N That's ok , Each line of ancient poetry is given in pinyin , Divide the sentence into two parts , Comma , Separate , Full stop . ending . A space is used to separate the Pinyin of two adjacent characters . The title ensures that the Pinyin of each word does not exceed 6 Characters , The total length of each line of characters does not exceed 100, also The second half of the poem has at least 3 A word .

That is to say, the first half of the sentence may have no words , So you need this code

if (pos < 3) {
    
			cout << "Skipped";
			if (i != N - 1)	cout << endl;
			continue;
		}

Otherwise, a test point will be lost

After special judgment

When it comes to length , Pay attention to whether the subscript of the target object will exceed the limit ！