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

算法分析：本题的逻辑比较简单，没有什么坑，按照题目描述的步骤来即可，只要注意输出单数和复数有区别就行了。由于要顺序输出，所以用了一个队列。

#include <stdio.h>#include <stdlib.h>#include <string.h>struct User{    char name[11];    char key[11];}users[1010];int main(){    int i, n, m = 0;    int queue[1010] = {0}, front = -1, tail = -1;    scanf("%d", &n);    for(i = 0; i < n; i++)    {        int j, f = 0;        scanf("%s %s", users[i].name, users[i].key);        for(j = 0; j < strlen(users[i].key); j++)        {            switch(users[i].key[j])            {                case '1': users[i].key[j] = '@'; break;                case '0': users[i].key[j] = '%'; break;                case 'l': users[i].key[j] = 'L'; break;                case 'O': users[i].key[j] = 'o'; break;                default : f++;            }        }        if(f < strlen(users[i].key)) //当改密码存在修改时        {            queue[++front] = i; //该密码所对应的用户进入队列            m++;         }    }    if(m > 0) //当一个用例中有用户的密码被修改    {        printf("%d\n", m);        while(front != tail) //出队        {            tail++;            printf("%s %s\n", users[queue[tail]].name, users[queue[tail]].key);        }    }    else    {        if(n == 1) //单复数分开输出            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;}