PAT-A1035 Password 题目内容及题解

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#PAT #甲级 #OJ #编程学习

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该博客主要讨论PAT甲级考试中关于Password检查的题目，内容包括输入和输出规格说明，以及样例输入和输出。博主解释了如何识别和解决数字与字母混淆的问题，例如将1替换为@，0替换为%，l替换为L，O替换为o。博主提供了解题思路，遍历密码并替换可能引起混淆的字符，同时记录了需要修改的账户数量。在文章末尾，博主给出了代码实现和运行结果。

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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

题目大意

按照题目要求替换字符并输出结果。

解题思路

遍历并替换字符，并记录有替换的记录数及记录编号；
按要求输出并返回0值。

代码

#include<stdio.h>
#include<string.h>
#define maxn 1010

struct Password{
    char id[15],pass[15];
}PW[maxn];

int main(){
    int c[maxn];
    int cnum=0;
    int len,i,j,flag;
    int N;
    scanf("%d",&N);
    for(i=0;i<N;i++){
        getchar();
        scanf("%s %s",&PW[i].id,&PW[i].pass);
        flag=0;
        len=strlen(PW[i].pass);
        for(j=0;j<len;j++){
            switch(PW[i].pass[j]){
                case '1':{
                    PW[i].pass[j]='@';
                    flag=1;
                    break;
                }
                case '0':{
                    PW[i].pass[j]='%';
                    flag=1;
                    break;
                }
                case 'l':{
                    PW[i].pass[j]='L';
                    flag=1;
                    break;
                }
                case 'O':{
                    PW[i].pass[j]='o';
                    flag=1;
                    break;
                }
            }
        }
        if(flag){
            c[cnum++]=i;
        }
    }
    if(cnum==0){
        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);
        }
    }else{
        printf("%d\n",cnum);
        for(i=0;i<cnum;i++){
            printf("%s %s\n",PW[c[i]].id,PW[c[i]].pass);
        }
    }
    return 0;
}