【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;
}