L1-059 敲笨钟

 PTA | 程序设计类实验辅助教学平台

 一开始我不明白，怎么把一句话分割输入，并且判断

下面是先输入在判断，如果逗号前面时候ong并且.前面也是ong那就是押韵的

并且根据空格判断第几个单词

#include <iostream>
using namespace std;

typedef long long ll;
const int maxn = 1e5 + 10;
string s;

int main() {
	int n;
	cin >> n;
	getchar();
	while (n--) {
		getline(cin, s);
		int f1 = 0, f2 = 0, num = 0;
		for (int i = 0; i < s.size(); ++i) {
			if (s[i] == ',')
				if (s[i - 3] == 'o' && s[i - 2] == 'n' && s[i - 1] == 'g')
					f1 = 1;
			if (s[i] == '.')
				if (s[i - 3] == 'o' && s[i - 2] == 'n' && s[i - 1] == 'g')
					f2 = 1;
			if (s[i] == ' ')
				num++;
		}
		if (f1 && f2) {
			for (int i = 0; i < s.size(); ++i) {
				if (num == 2)
					break;
				cout << s[i];
				if (s[i] == ' ')
					num--;
			}
			cout << "qiao ben zhong.";
		} else {
			cout << "Skipped";
		}
		cout << "\n";
	}
	return 0;
}

下面判断有点不同，他是分成两个输入，把getline玩明白了属于是

然后倒序判断该在哪里输出

#include <iostream>
using namespace std;

int main() {
	int n;
	cin >> n;
	while (n--) {
		cin.get(); // 和getchar效果一样
		string s1, s2;
		getline(cin, s1, ',');
		getline(cin, s2, '.');
		
		int l1 = s1.size(), l2 = s2.size();
		
		if (s1[l1 - 1] != 'g' || s1[l1 - 2] != 'n' || s1[l1 - 3] != 'o' || s2[l2 - 1] != 'g' || s2[l2 - 2] != 'n'
		        || s2[l2 - 3] != 'o')
			puts("Skipped");
		else {
			cout << s1 << ',';
			int i, cnt = 0;
			for (i = s2.size(); i >= 0; i--) {
				if (s2[i] == ' ')
					cnt++;
				if (cnt == 3)
					break;
			}
			for (int j = 0; j <= i; j++)
				cout << s2[j];
			cout << "qiao ben zhong.\n";
		}
	}

	return 0;
}