浙大pat 1035题解
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
#include"iostream"
#include "algorithm"
#include "string"
#include "vector"
using namespace std;
struct PAT
{
string username;
string password;
};
int num =0;
bool FindWords(string &s)
{
bool temp = false;
int n = s.size();
int i;
for( i=0;i<n;i++)
{
if(s[i]=='1')
{
s[i]='@';
temp =true;
}
if(s[i]=='0')
{
s[i]='%';
temp =true;
}
if(s[i]=='l')
{
s[i]='L';
temp =true;
}
if(s[i]=='O')
{
s[i]='o';
temp =true;
}
}
return temp;
}
int main()
{
vector<PAT>p;
int n;
cin >> n;
PAT pat;
for(int i=0;i<n;i++)
{
cin >> pat.username >> pat.password;
if(FindWords(pat.password))
{
p.push_back(pat);
num++;
}
}
if(n==1&&num==0)
cout<<"There is 1 account and no account is modified"<<endl;
else if(num==0)
cout<<"There are "<<n<<" accounts and no account is modified"<<endl;
else if(num!=0)
{
cout << num<<endl;
vector<PAT>::iterator it = p.begin();
while(it!=p.end())
{
cout << (*it).username<<" "<<(*it).password<<endl;
it++;
}
}
return 0;
}
浙大pat 1035题解的更多相关文章
- 浙大pat 1025题解
1025. PAT Ranking (25) 时间限制 200 ms 内存限制 32000 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yue Programmi ...
- 浙大pat 1011题解
With the 2010 FIFA World Cup running, football fans the world over were becoming increasingly excite ...
- 浙大PAT 7-06 题解
#include <stdio.h> #include <iostream> #include <algorithm> #include <math.h> ...
- 浙大pat 1012题解
1012. The Best Rank (25) 时间限制 400 ms 内存限制 32000 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yue To eval ...
- 浙大 pat 1003 题解
1003. Emergency (25) 时间限制 400 ms 内存限制 32000 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yue As an emerg ...
- 浙大 pat 1038 题解
1038. Recover the Smallest Number (30) 时间限制 400 ms 内存限制 32000 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHE ...
- 浙大 pat 1047题解
1047. Student List for Course (25) 时间限制 400 ms 内存限制 64000 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Y ...
- 浙大pat 1054 题解
1054. The Dominant Color (20) 时间限制 100 ms 内存限制 32000 kB 代码长度限制 16000 B 判题程序 Standard Behind the scen ...
- 浙大pat 1059 题解
1059. Prime Factors (25) 时间限制 50 ms 内存限制 32000 kB 代码长度限制 16000 B 判题程序 Standard 作者 HE, Qinming Given ...
随机推荐
- 【LeetCode】29. Divide Two Integers
题意:不用乘除求余运算,计算除法,溢出返回INT_MAX. 首先考虑边界条件,什么条件下会产生溢出?只有一种情况,即返回值为INT_MAX+1的时候. 不用乘除求余怎么做? 一.利用减法. 耗时太长, ...
- bootstrap 混合标签
<html lang="zh_cn"> <head> <meta charset="utf-8"> <meta htt ...
- 将mysql的data目录移走方法
如移动到"/home/mysql/data",我的mysql是装在/usr/local/mysql下的 1. 将/usr/local/mysql/data移动到/home/mysq ...
- WPF和Winform的一些界面控件
DevExpressTelerikMahApps.MetroModern UI for WPFModernWPFExtended WPF Toolkit™ Community EditionModer ...
- 两句话帮你彻底记住gdb之eXamining memory
对于刚学习Unix/Linux环境C编程的小朋友们或者写了很多所谓的C代码的老手们(其实很可能是机械程序员或者是伪程序员)来说,要记住gdb的eXaming memory的语法其实是相当不容易的,如果 ...
- Java泛型的定义以及对于<? extends T>和<? super T>
Java 中对于泛型方法的定义: public <T> T getT(){ .....相关代码; } 其中我对<T>的理解就是申明只是一个泛型方法的标记,T是返回的类型. 对于 ...
- 编译C语言单元测试框架CUnit库的方法
引用: http://blog.csdn.net/yygydjkthh/article/details/46357421 个人备忘使用 /******************************* ...
- metasploit nessus & db_autopwn
nessus官网:https://www.tenable.com/products/nessus-vulnerability-scanner 下载地址:https://www.tenable.com/ ...
- 凭借5G研究优势,诺基亚将携手菲律宾将其应用于VR/AR领域
目前,很多人都在抱怨网速不行,影响视频的流畅播放,未来这些问题可以通过5G解决.近日,诺基亚和PLDT的全资子公司Smart首次在菲律宾一个"现场"网络演示上实现了5G速度,该网络 ...
- [转]Mac常用软件推荐
https://github.com/hzlzh/Best-App