1035. Password (20)-PAT甲级真题

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：给定n个用户的姓名和密码，把密码中的1改为@，0改为%，l改为L，O改为o
如果不存在需要修改的密码，则输出There are n accounts and no account is modified。注意单复数，如果只有一个账户，就输出There is 1 account and no account is modified

分析：把需要改变的字符串改变后存储在字符串数组vector里面，根据数组里面元素的个数是否为0输出相应的结果

#include <iostream>
#include <vector>
using namespace std;
int main() {
    int n;
    scanf("%d", &n);
    vector<string> v;
    for(int i = 0; i < n; i++) {
        string name, s;
        cin >> name >> s;
        int len = s.length(), flag = 0;
        for(int j = 0; j < len; j++) {
            switch(s[j]) {
                case '1' : s[j] = '@'; flag = 1; break;
                case '0' : s[j] = '%'; flag = 1; break;
                case 'l' : s[j] = 'L'; flag = 1; break;
                case 'O' : s[j] = 'o'; flag = 1; break;
            }
        }
        if(flag) {
            string temp = name + " " + s;
            v.push_back(temp);
        }
    }
    int cnt = v.size();
    if(cnt != 0) {
        printf("%d\n", cnt);
        for(int i = 0; i < cnt; i++)
            cout << v[i] << endl;
    } else if(n == 1) {
        printf("There is 1 account and no account is modified");
    } else {
        printf("There are %d accounts and no account is modified", n);
    }
    return 0;
}

#include <iostream>

#include <vector>

using namespace std;

int main() {

int n;

scanf("%d", &n);

vector<string> v;

for(int i = 0; i < n; i++) {

string name, s;

cin >> name >> s;

int len = s.length(), flag = 0;

for(int j = 0; j < len; j++) {

switch(s[j]) {

case '1' : s[j] = '@'; flag = 1; break;

case '0' : s[j] = '%'; flag = 1; break;

case 'l' : s[j] = 'L'; flag = 1; break;

case 'O' : s[j] = 'o'; flag = 1; break;

}

if(flag) {

string temp = name + " " + s;

v.push_back(temp);

}

int cnt = v.size();

if(cnt != 0) {

printf("%d\n", cnt);

for(int i = 0; i < cnt; i++)

cout << v[i] << endl;

} else if(n == 1) {

printf("There is 1 account and no account is modified");

} else {

printf("There are %d accounts and no account is modified", n);

}

return 0;

}

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