Educational Codeforces Round 78 (Rated for Div. 2) A. Shuffle Hashing
链接:
https://codeforces.com/contest/1278/problem/A
题意:
Polycarp has built his own web service. Being a modern web service it includes login feature. And that always implies password security problems.
Polycarp decided to store the hash of the password, generated by the following algorithm:
take the password p, consisting of lowercase Latin letters, and shuffle the letters randomly in it to obtain p′ (p′ can still be equal to p);
generate two random strings, consisting of lowercase Latin letters, s1 and s2 (any of these strings can be empty);
the resulting hash h=s1+p′+s2, where addition is string concatenation.
For example, let the password p= "abacaba". Then p′ can be equal to "aabcaab". Random strings s1= "zyx" and s2= "kjh". Then h= "zyxaabcaabkjh".
Note that no letters could be deleted or added to p to obtain p′, only the order could be changed.
Now Polycarp asks you to help him to implement the password check module. Given the password p and the hash h, check that h can be the hash for the password p.
Your program should answer t independent test cases.
思路:
暴力枚举
代码:
#include<bits/stdc++.h>
using namespace std;
map<char, int> Mp;
string p, s;
bool Check(int x)
{
map<char, int> Tmp;
Tmp = Mp;
for (int i = 0;i < (int)p.size();i++)
{
if (Tmp[s[i+x]] < 0)
return false;
Tmp[s[i+x]]--;
}
for (auto v: Tmp) if (v.second > 0)
return false;
return true;
}
int main()
{
int t;
cin >> t;
while(t--)
{
Mp.clear();
cin >> p >> s;
for (int i = 0;i < (int)p.size();i++)
Mp[p[i]]++;
bool flag = false;
for (int i = 0;i < (int)s.size();i++) if (Check(i))
{
flag = true;
break;
}
if (flag)
cout << "YES" << endl;
else
cout << "NO" << endl;
}
return 0;
}
Educational Codeforces Round 78 (Rated for Div. 2) A. Shuffle Hashing的更多相关文章
- 【cf比赛记录】Educational Codeforces Round 78 (Rated for Div. 2)
比赛传送门 A. Shuffle Hashing 题意:加密字符串.可以把字符串的字母打乱后再从前面以及后面接上字符串.问加密后的字符串是否符合加密规则. 题解:字符串的长度很短,直接暴力搜索所有情况 ...
- Educational Codeforces Round 78 (Rated for Div. 2) D. Segment Tree
链接: https://codeforces.com/contest/1278/problem/D 题意: As the name of the task implies, you are asked ...
- Educational Codeforces Round 78 (Rated for Div. 2) C. Berry Jam
链接: https://codeforces.com/contest/1278/problem/C 题意: Karlsson has recently discovered a huge stock ...
- Educational Codeforces Round 78 (Rated for Div. 2) B. A and B
链接: https://codeforces.com/contest/1278/problem/B 题意: You are given two integers a and b. You can pe ...
- Educational Codeforces Round 78 (Rated for Div. 2)B. A and B(1~n的分配)
题:https://codeforces.com/contest/1278/problem/B 思路:还是把1~n分配给俩个数,让他们最终相等 假设刚开始两个数字相等,然后一个数字向前走了abs(b- ...
- Educational Codeforces Round 78 (Rated for Div. 2)
A题 给出n对串,求s1,是否为s2一段连续子串的重排,串长度只有100,从第一个字符开始枚举,sort之后比较一遍就可以了: char s1[200],s2[200],s3[200]; int ma ...
- Educational Codeforces Round 78 (Rated for Div. 2) --补题
链接 直接用数组记录每个字母的个数即可 #include<bits/stdc++.h> using namespace std; int a[26] = {0}; int b[26] = ...
- Educational Codeforces Round 78 (Rated for Div. 2) 题解
Shuffle Hashing A and B Berry Jam Segment Tree Tests for problem D Cards Shuffle Hashing \[ Time Lim ...
- Educational Codeforces Round 78 (Rated for Div. 2) C - Berry Jam(前缀和)
随机推荐
- 强迫症福利--收起.NET程序的dll来
作为上床后需要下床检查好几次门关了没有的资深强迫症患者,有一个及其搞我的问题,就是dll问题. 曾几何时,在没有nuget的年代,当有依赖项需要引用的时候,只能通过文件引用来管理引用问题,版本问题,更 ...
- C#程序只允许运行一个实例的解决方案
最近在做winform的程序中,需要只能打开一个程序,如果已经存在,则激活该程序的窗口,并显示在最前端.在网上google了一哈,找到了很多的解决方案.这里我整理了3种方案,并经过了测试,现和朋友们分 ...
- 使用maven-resources-plugin插件分环境配置
一.项目目录结构 二.pom文件中引入maven-resources-plugin插件和相关的标签 <build> <plugins> <plugin> &l ...
- SpringBoot之CommandLineRunner接口和ApplicationRunner接口
我们在开发中可能会有这样的情景.需要在容器启动的时候执行一些内容.比如读取配置文件,数据库连接之类的.SpringBoot给我们提供了两个接口来帮助我们实现这种需求.这两个接口分别为CommandLi ...
- jq数字翻页效果,随机数字显示,实现上下翻动效果
最近在做一个项目,需要实时展示一串数字,要有类似于日历翻页的效果,在网上找寻了一番,发现dataStatistics这个插件http://www.jq22.com/jquery-info8141能实现 ...
- markdown文本内跳转
Markdown文本内跳转 构建茅的过程中使用markdown语法,类似于markdown向外跳转链接,目的地地址写成#. 在markdown文本中写入: 目录 跳转 跳转部分按照html文本的写法 ...
- HTTP之URL的快捷方式
URL快捷方式 ==================摘自<HTTP权威指南>======================= WEB客户端可以理解并使用几种URL快捷方式.相对URL是在某职 ...
- 前端开发vscode必备插件
VSCode 插件 Atom one Dark Theme Atom Dark主题 Auto Close Tag 自动关闭标签 Auto Rename Tag 自动重命名标签 Beautify 格式化 ...
- MongoDB学习笔记(六)
初识 MongoDB 中的索引 索引就像图书的目录一样,可以让我们快速定位到需要的内容,关系型数据库中有索引,NoSQL 中当然也有,本文我们就先来简单介绍下 MongoDB 中的索引. 索引创建 默 ...
- ElasticSearch如何更新集群的状态
ElasticSearch如何更新集群的状态 最近发生了很多事情,甚至对自己的技术能力和学习方式产生了怀疑,所以有一段时间没更新文章了,估计以后更新的频率会越来越少,希望有更多的沉淀而不是简单地分享. ...