1035 Password (20分)

通过代码：

#include <iostream>

using namespace std;

const int N = 1010;

string name[N], pwd[N];

string change(string str)
{
    string res;

    for (auto c : str)  //范围遍历
        if (c == '1') res += '@';
        else if (c == '0') res += '%';
        else if (c == 'l') res += 'L';
        else if (c == 'O') res += 'o';
        else res += c;

    return res;
}

int main()
{
    int n;
    cin >> n;

    int m = 0;
    for (int i = 0; i < n; i ++ )
    {
        string cur_name, cur_pwd;
        cin >> cur_name >> cur_pwd;
        string changed_pwd = change(cur_pwd);

        if (changed_pwd != cur_pwd)
        {
            name[m] = cur_name; //修改过密码的姓名
            pwd[m] = changed_pwd; //修改后的密码
            m ++ ;  // 计数
        }
    }

    if (!m)
    {
        if (n == 1) puts("There is 1 account and no account is modified");
        else printf("There are %d accounts and no account is modified\n", n);
    }
    else
    {
        cout << m << endl;
        for (int i = 0; i < m; i ++ ) cout << name[i] << ' ' << pwd[i] << endl;
    }

    return 0;
}

思路：

1. 更新密码之后于原密码比较，查看是否一样
2. 单独写一个change函数
3. 在没有修改密码的情况下，特判n=0的情况