1035 Password(20)

To prepare for PAT, the judge sometimes has to generate random passwords for the users. The problem is that there are always some confusing passwords since it is hard to distinguish 1 (one) from l (L in lowercase), or 0 (zero) from O (o in uppercase). One solution is to replace 1 (one) by @, 0 (zero) by %, l by L, and O by o. Now it is your job to write a program to check the accounts generated by the judge, and to help the juge modify the confusing passwords.

Input Specification:

Each input file contains one test case. Each case contains a positive integer N (≤1000), followed by N lines of accounts. Each account consists of a user name and a password, both are strings of no more than 10 characters with no space.

Output Specification:

For each test case, first print the number M of accounts that have been modified, then print in the following M lines the modified accounts info, that is, the user names and the corresponding modified passwords. The accounts must be printed in the same order as they are read in. If no account is modified, print in one line There are N accounts and no account is modified where N is the total number of accounts. However, if N is one, you must print There is 1 account and no account is modified instead.

Sample Input 1:

3
Team000002 Rlsp0dfa
Team000003 perfectpwd
Team000001 R1spOdfa

Sample Output 1:

2
Team000002 RLsp%dfa
Team000001 R@spodfa

Sample Input 2:

1
team110 abcdefg332

Sample Output 2:

There is 1 account and no account is modified

Sample Input 3:

2
team110 abcdefg222
team220 abcdefg333

Sample Output 3:

There are 2 accounts and no account is modified

Solution:

题目要求出修改密码后的信息.

1类分成, 没被修改的, 修改的;

2类没被修改的分成, 复数和单数.

把修改的字符建成unodered_map用于查找.
用struct将name和password关联, 读入password进行修改和过滤, 被修改的node放入vector;
vector.size(), N决定了属于上述哪一类。

#include <cstdio>
#include <cstring>
#include <vector>
#include <unordered_map>
using namespace std;

struct Node {
  char name[11];
  char password[11];
};

int main() {
  int N; scanf("%d", &N);
  unordered_map<char, char> mp;
  mp['1'] = '@'; mp['0'] = '%'; mp['l'] = 'L'; mp['O'] = 'o';
  vector<Node> v;
  for (int i = 0; i < N; i++) {
    Node node;
    scanf("%s %s", node.name, node.password);
    int modified = 0;
    for (int j = 0; j < strlen(node.password); j++) {
      auto it = mp.find(node.password[j]);
      if (it != mp.end()) {
        node.password[j] = it->second;
        modified = 1;
      }
    }
    if (modified) {
      v.push_back(node);
    }
  }

  if (v.size() != 0) {
    printf("%lu\n", v.size());
    for (int i = 0; i < v.size(); i++) {
      printf("%s %s\n", v[i].name, v[i].password);
    }
  } else {
    if (N == 1) {
      printf("There is 1 account and no account is modified\n");
    } else {
      printf("There are %d accounts and no account is modified\n", N);
    }
  }

  return 0;
}