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) froml
(L
in lowercase), or0
(zero) fromO
(o
in uppercase). One solution is to replace1
(one) by@
,0
(zero) by%
,l
byL
, andO
byo
. 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
whereN
is the total number of accounts.However, if
N
is one, you must printThere 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
0---->% 1------->@ l------>L O-------> o
Do not modify the output
#include<iostream>
#include<vector>
using namespace std;
int main()
{
int n;
cin>>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 '0':s[j]='%';flag=1;break;
case '1':s[j]='@';flag=1;break;
case 'l':s[j]='L';flag=1;break;
case 'O':s[j]='o';flag=1;break;
}
}///替换所有字符
if(flag!=0)
{
string temp=name+" "+s;
v.push_back(temp);
}
}///把改造完成的字符串传进vector数组中
int cnt=v.size();///判断是否为空
if(cnt!=0)
{
cout<<cnt<<endl;
for(int i=0;i<cnt;i++)
cout<<v[i]<<endl;
}///vector中的一个元素v[i]为name+" "+s的格式
else if(n==1)
cout<<"There is 1 account and no account is modified";
else
cout<<"There are "<<n<<" accounts and no account is modified";
}