PAT-A 1035. Password (20)

传送门

https://www.patest.cn/contests/pat-a-practise/1035

题目

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

分析

开始是用map来做的,但是发现map还帮着给学号排序了,于是后来选用了vector,并且在vector中放pair对象。

如果密码不需要修改的话,不需要记录密码,若密码被修改了,则存入vector中。

最后如果所有账户都未被修改,输出题目给定的那句话,记得这里要区分单复数的问题。
单数的话是:There is 1 account and no account is modified
复数的话是:There are 2 accounts and no account is modified
这道题比较良心,给出单复数的例子了,记得乙级有道题,连例子都没给,坑了一大批人。

源代码

#include <iostream>
#include <string>
#include <vector>

using namespace std;

int main(){
    int n;
    scanf("%d", &n);
    string no, pwd;
    int count = 0;
    vector<pair<string, string> > v;
    for(int i = 0; i < n; ++i){
        cin >> no >> pwd;
        bool flag = true;
        for(int i = 0; i < pwd.size(); ++i){
            if(pwd[i] == '1'){
                pwd[i] = '@';
                flag = false;
            }
            else if(pwd[i] == '0'){
                pwd[i] = '%';
                flag = false;
            }
            else if(pwd[i] == 'l'){
                pwd[i] = 'L';
                flag = false;
            }
            else if(pwd[i] == 'O'){
                pwd[i] = 'o';
                flag = false;
            }
        }
        if(flag){
            ++count; //有一个账户未被修改
        }
        else{
            pair<string, string> p(no, pwd);
            v.push_back(p);
        }
    }
    if(count == n){
        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", v.size());
        for(int i = 0; i < v.size(); ++i){
            cout << v[i].first << " " << v[i].second << endl;
        }
    }
    return 0;
}

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