1035 Password (20分)(水)
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 (≤), 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
题目分析:水题 细心就好
#define _CRT_SECURE_NO_WARNINGS
#include <climits>
#include<iostream>
#include<vector>
#include<queue>
#include<map>
#include<stack>
#include<algorithm>
#include<string>
#include<cmath>
using namespace std;
struct S {
string s1, s2;
};
vector<S> V;
int main()
{
int N;
cin >> N;
string s1, s2;
for (int i = ; i < N; i++)
{
int flag = ;
cin >> s1 >> s2;
for (int j = ; j < s2.length(); j++)
{
if (s2[j] == '')
{
flag = ;
s2[j] = '@';
}
else if (s2[j] == '')
{
flag = ;
s2[j] = '%';
}
else if (s2[j] == 'l')
{
flag = ;
s2[j] = 'L';
}
else if (s2[j] == 'O')
{
flag = ;
s2[j] ='o';
}
}
if (flag)
V.push_back({ s1,s2 });
}
if (V.size() == )
{
if (N == )
cout << "There is 1 account and no account is modified";
else
printf("There are %d accounts and no account is modified", N);
}
else
{
cout << V.size() << endl;
for (auto it : V)
cout << it.s1 << " " << it.s2<<endl;
}
}
1035 Password (20分)(水)的更多相关文章
- PAT 甲级 1035 Password (20 分)(简单题)
1035 Password (20 分) To prepare for PAT, the judge sometimes has to generate random passwords for ...
- PAT (Advanced Level) Practice 1035 Password (20 分) 凌宸1642
PAT (Advanced Level) Practice 1035 Password (20 分) 凌宸1642 题目描述: To prepare for PAT, the judge someti ...
- PAT甲级——1035 Password (20分)
To prepare for PAT, the judge sometimes has to generate random passwords for the users. The problem ...
- PAT Advanced 1035 Password (20 分)
To prepare for PAT, the judge sometimes has to generate random passwords for the users. The problem ...
- PAT (Advanced Level) Practice 1035 Password (20 分)
To prepare for PAT, the judge sometimes has to generate random passwords for the users. The problem ...
- 【PAT甲级】1035 Password (20 分)
题意: 输入一个正整数N(<=1000),接着输入N行数据,每行包括一个ID和一个密码,长度不超过10的字符串,如果有歧义字符就将其修改.输出修改过多少组密码并按输入顺序输出ID和修改后的密码, ...
- PAT甲题题解-1035. Password (20)-水
题意:给n个用户名和密码,把密码中的1改为@,0改为%,l改为L,O改为o. 让你输出需要修改密码的用户名个数,以及对应的用户名和密码,按输入的顺序.如果没有用户需要修改,则输出对应的语句,注意单复数 ...
- 【PAT】1035. Password (20)
题目:http://pat.zju.edu.cn/contests/pat-a-practise/1035 分析:简单题.直接搜索,然后替换,不会超时,但是应该有更好的办法. 题目描述: To pre ...
- 1035 Password (20)
#include <stdio.h> #include <string.h> struct MyStruct { ]; ]; bool changed; }; int main ...
随机推荐
- Redis系列二 - 数据结构
前言 redis作为我们开发的一大神器,我们接触肯定不会少,但是很多同学也许只会存储String类型的值,这是非常不合理的.在这里,将带大家认识Redis的5中数据结构. 1.问:Redis有那些数据 ...
- 【vue】---- v-model在自定义组件中的使用
1. v-model简介 可以用 v-model 指令在表单 <input>.<textarea> 及 <select> 元素上创建双向数据绑定,它的本质是一个语法 ...
- 《ASP.NET Core 3框架揭秘》博文汇总
在过去一段时间内,写了一系列关于ASP.NET Core 3相关的文章,其中绝大部分来源于即将出版的<ASP.NET Core 3框架揭秘>(博文只能算是"初稿",与书 ...
- 利用border-radius画椭圆
<!DOCTYPE html> <html> <head> <meta charset="utf-8"> <title> ...
- P5596 【XR-4】题 笔记
P5596 [XR-4]题 其实这题我昨天没做出来--所以今天写一下笔记 昨天我还信誓旦旦地说这一定是一道黑题\(OTZ\).果然菜是原罪. 另外吐槽一下科技楼机房频繁停电,昨天写了两小时的树刨和倍增 ...
- 判断 tableZen 是否有 横向滚动条
判断 tableZen 是否有 横向滚动条 const outWidth = this.$refs.tableInnerZen.$el.clientWidth ].$el.clientWidth
- 14. LiveBos编号自动生成
(1) var temp="Apex"; var no=""+ABS_DYNSERIALNO(true,temp); var len=no.length; va ...
- 使用tomcat运行时提示some characters cannot be mapped using iso-8859-1 character encoding异常
今天第一次使用java进行jsp项目搭建,也是第一次使用tomcat.tomcat是运行java web的一个小型服务器,属于Apache的一个开源免费的服务. 在运行web 的时候,我们就要先配置好 ...
- 【攻防世界】simple-unpack
知识:upx脱壳 simple-unpack 难度系数: 3.0 题目来源: 暂无 题目描述:菜鸡拿到了一个被加壳的二进制文件 提示说有壳子:然后用PE分析发现是ELF upx的壳子
- [深入学习C#] 匿名函数、委托和Lambda表达式
匿名函数 匿名函数(Anonymous Function)是表示“内联”方法定义的表达式.匿名函数本身及其内部没有值或者类型,但是可以转换为兼容的委托或者表达式树类型(了解详情).匿名函数转换的计算取 ...