PAT日志 1035

最新推荐文章于 2021-07-09 09:53:40 发布

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本文介绍了一个编程挑战，旨在解决密码中容易混淆的字符问题。通过将1替换为@，0替换为%，l替换为L，O替换为o，来增强密码的可读性和区分度。文章提供了输入输出规范及样例，同时强调了单数和复数输出的区别。

顽强的小白

1035 Password （20 分）

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

题目解析

题目很简单就是把密码中的四个字符改成相应的字符来进行区分，易错点就是细节了。
要注意的是，如果没有更改的话，有两种输出方式，一种是只有一个数据的情况，注意看例子中的输出句子是单数的！另一种是复数输出。

代码实现

 char id[20];
 char pw[20];
 int tag;
};
int main(){
 int n,cnt=0;
 student stu[1005];
 for(int i=0;i<1005;i++) stu[i].tag=0;
 scanf("%d",&n);
 for(int i=0;i<n;++i ){
  scanf("%s %s",stu[i].id,stu[i].pw);
  int len=strlen(stu[i].pw);
  int flag=0;
  for(int j=0;j<len;++j){
   if(stu[i].pw[j]=='1'){
    stu[i].pw[j]='@';
    stu[i].tag=1;
    flag=1;
   }else if(stu[i].pw[j]=='0'){
    stu[i].pw[j]='%';
    stu[i].tag=1;
    flag=1;
   }else if(stu[i].pw[j]=='l'){
    stu[i].pw[j]='L';
    stu[i].tag=1;
    flag=1;
   }else if(stu[i].pw[j]=='O'){
    stu[i].pw[j]='o';
    stu[i].tag=1; 
    flag=1;
   } 
  
  }
  if(flag)++cnt;
 }
 if(cnt==0&&n==1){
  printf("There is 1 account and no account is modified");
  return 0;
 }else if(cnt==0){
  printf("There are %d accounts and no account is modified",n);
  return 0;
 }
 printf("%d\n",cnt);
 for(int i=0;i<n;++i){
  if(stu[i].tag){
   printf("%s %s\n",stu[i].id,stu[i].pw);
  }
 }
 
}