1035 Password
part 1, 1.1
自己解法
- 使用Account类，看起来比较麻烦，但是逻辑清晰
#include <iostream>
using namespace std;
#include <string>
#include <vector>
#include <algorithm>class Account
{
public:string username;string password;Account(string username, string password){this->username = username;for (int i = 0; i < password.length(); i++){switch (password[i]){case '1':password[i] = '@';break;case '0':password[i] = '%';break;case 'l':password[i] = 'L';break;case 'O':password[i] = 'o';break;}}this->password = password;}
};class printModified
{
public:void operator()(Account a){cout << a.username << " " << a.password << endl;}
};int main()
{int n;cin >> n;int m = n, temp = n;vector<Account> v;while (n--){string username;string password;cin >> username >> password;if (password.find("1") != string::npos || password.find("0") != string::npos || password.find("l") != string::npos || password.find("O") != string::npos){Account a(username, password);v.push_back(a);m--;}}if (m == temp){if (temp == 1)cout << "There is " << m << " account and no account is modified" << endl;elsecout << "There are " << m << " accounts and no account is modified" << endl;}else{cout << temp - m << endl;for_each(v.begin(), v.end(), printModified());}system("pause");return 0;
}
大神解法
- 柳神
- 将所有修改好的 name + password 放到 v 中，也很清晰
#include <iostream>
#include <vector>
using namespace std;
int main() {int n;scanf("%d", &n);vector<string> v;for(int i = 0; i < n; i++) {string name, s;cin >> name >> s;int len = s.length(), flag = 0;for(int j = 0; j < len; j++) {switch(s[j]) {case '1' : s[j] = '@'; flag = 1; break;case '0' : s[j] = '%'; flag = 1; break;case 'l' : s[j] = 'L'; flag = 1; break;case 'O' : s[j] = 'o'; flag = 1; break;}}if(flag) {string temp = name + " " + s;v.push_back(temp);}}int cnt = v.size();if(cnt != 0) {printf("%d\n", cnt);for(int i = 0; i < cnt; i++)cout << v[i] << endl;} else if(n == 1) {printf("There is 1 account and no account is modified");} else {printf("There are %d accounts and no account is modified", n);}return 0;
}
总结
- 注意有个 is 和 are 的区别，以及account单复数(OUTPUT2 和 OUTPUT3)????
- string.find() 如果没找到，返回 string::npos