A1035 Password (20)（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 [email protected]

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

思考

c++的引用很方便，传递参数，传递了操作对象本身，灵魂附体。

c语言没有引用，只能指针取地址。

AC代码

c语言

#include <stdio.h>
#include <string.h>
#include <stdbool.h>
struct node {
    char name[20], password[20];
    bool ischange;
}T[1005];
/*在C语言中是不存在引用的，也就是说C语言中&表示的不是引用，仅仅是取地址符。所以错误提示就是告诉你&在这里用的不对，那怎么解决呢？

首先介绍一个正规的解决方法：用指针来取代引用，在主函数中传进来地址；*/
void crypt(struct node* t, int* cnt) {//c语言结构体必须加struct,typedef可解决此类问题
    int len = strlen(t->password);
    for(int i = 0; i < len; i++) {
        if(t->password[i] == ‘1‘) {
            t->password[i] = ‘@‘;
            t->ischange = true;
        } else if(t->password[i] == ‘0‘) {
            t->password[i] = ‘%‘;
            t->ischange = true;
        } else if(t->password[i] == ‘l‘) {
            t->password[i] = ‘L‘;
            t->ischange = true;
        } else if(t->password[i] == ‘O‘) {
            t->password[i] = ‘o‘;
            t->ischange = true;
        }
    }
    if(t->ischange) {
        (*cnt)++;//优先级很重要，先解指针，再自增1，遇到优先级没把握，加括号，都加上
    }
}

int main() {
    int n, cnt = 0;
    scanf("%d", &n);
    for(int i = 0; i < n; i++) {
        scanf("%s %s", T[i].name, T[i].password);
        T[i].ischange = false;//初始化为未修改
    }
    for(int i = 0; i < n; i++) {
        crypt(&T[i], &cnt);
    }
    if(cnt == 0) {
        if(n == 1) printf("There is %d account and no account is modified", n);
        else {
            printf("There are %d accounts and no account is modified", n);
        }
    }else {
        printf("%d\n", cnt);
        for(int i = 0; i < n; i++) {
            if(T[i].ischange) {
                printf("%s %s\n", T[i].name, T[i].password);
            }
        }
    }
    return 0;
}

c++

#include <cstdio>
#include <cstring>
struct node {
    char name[20], password[20];
    bool ischange;
}T[1005];

void crypt(node& t, int& cnt) {//引用，可以对传入参数进行修改
    int len = strlen(t.password);
    for(int i = 0; i < len; i++) {
        if(t.password[i] == ‘1‘) {
            t.password[i] = ‘@‘;
            t.ischange = true;
        } else if(t.password[i] == ‘0‘) {
            t.password[i] = ‘%‘;
            t.ischange = true;
        } else if(t.password[i] == ‘l‘) {
            t.password[i] = ‘L‘;
            t.ischange = true;
        } else if(t.password[i] == ‘O‘) {
            t.password[i] = ‘o‘;
            t.ischange = true;
        }
    }
    if(t.ischange) {
        cnt++;
    }
}

int main() {
    int n, cnt = 0;
    scanf("%d", &n);
    for(int i = 0; i < n; i++) {
        scanf("%s %s", T[i].name, T[i].password);
        T[i].ischange = false;
    }
    for(int i = 0; i < n; i++) {
        crypt(T[i], cnt);
    }
    if(cnt == 0) {
        if(n == 1) printf("There is %d account and no account is modified", n);
        else {
            printf("There are %d accounts and no account is modified", n);
        }
    }else {
        printf("%d\n", cnt);
        for(int i = 0; i < n; i++) {
            if(T[i].ischange) {
                printf("%s %s\n", T[i].name, T[i].password);
            }
        }
    }
    return 0;
}

原文地址：https://www.cnblogs.com/lingr7/p/9452825.html