天天看點

OpenJudge NOI 1.7 22:緊急措施

【題目連結】

OpenJudge NOI 1.7 22:緊急措施

【題目考點】

1. 字元串

【解題思路】

先輸入自己的email和n,循環輸入n組資料,如果一條記錄的email和自己的email相同,那麼将這條記錄的密碼的字母大小寫轉換,輸出使用者名和新的密碼。

【題解代碼】

解法1:使用字元數組

#include <bits/stdc++.h>
using namespace std;
#define N 55
int main()
{
	char user[N], pwd[N], email[N], myEmail[N];
	bool hasOut = false;
	int n;
	cin >> myEmail >> n;
	for(int i = 1; i <= n; ++i)
	{
		cin >> user >> pwd >> email;
		if(strcmp(email, myEmail) == 0)
		{
		    int len = strlen(pwd); 
			for(int j = 0; j < len; ++j)
			{
				if(pwd[j] >= 'a' && pwd[j] <= 'z')
					pwd[j] -= 32;//小寫轉大寫 
				else if(pwd[j] >= 'A' && pwd[j] <= 'Z')
					pwd[j] += 32;//大寫轉小寫 
			}
			cout << user << ' ' << pwd << endl;
			hasOut = true;
		}
	}
	if(hasOut == false)
		cout << "empty";
	return 0;
}
           

解法2:使用string類與<cctype>函數

#include <bits/stdc++.h>
using namespace std;
int main()
{
	string user, pwd, email, myEmail;
	bool hasOut = false;
	int n;
	cin >> myEmail >> n;
	for(int i = 1; i <= n; ++i)
	{
		cin >> user >> pwd >> email;
		if(email == myEmail)
		{
			for(int j = 0; j < pwd.size(); ++j)
			{
				if(islower(pwd[j]))
					pwd[j] = toupper(pwd[j]);
				else if(isupper(pwd[j]))
					pwd[j] = tolower(pwd[j]);
			}
			cout << user << ' ' << pwd << endl;
			hasOut = true;
		}
	}
	if(hasOut == false)
		cout << "empty";
	return 0;
}
           

繼續閱讀