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

  1. 【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 ...

  2. 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 ...

  3. 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 ...

  4. 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 ...

  5. 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- ...

  6. 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 ...

  7. 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] = ...

  8. 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 ...

  9. Educational Codeforces Round 78 (Rated for Div. 2) C - Berry Jam(前缀和)

随机推荐

  1. 写了此书《微软Azure实战参考》 北京航空航天大学出版社 2019年7月

    经过大半年的努力,写出了此书<微软Azure实战参考> 北京航空航天大学出版社 2019年7月.喜爱微软云平台和需要使用微软云技术的读者,可以参考一下.

  2. 问题查询-tomcat内存泄露

    1.报警信息 内容: 微信服务器向公众号推送消息或事件后,开发者5秒内没有返回 次数: 5分钟 239次 错误样例: [OpenID=o][Stamp=1562718361][3rdUrl=url][ ...

  3. .NET Core:路由

    (1)模板路由 在Startup的Configure方法中配置: app.UseMvc(routes =>{ routes.MapRoute( name: "areas", ...

  4. PHP获取cookie、Token、模拟登录、抓取数据、解析生成json

    本文介绍使用PHP获取cookie,获取Token.以及模拟登录.然后抓取数据.最后解析生成json的的过程. 0. 设置Cookie路径 set_time_limit(0); //使用的cookie ...

  5. 1 datax 安装和简单使用

    DataX 是阿里巴巴集团内被广泛使用的离线数据同步工具/平台,实现包括 MySQL.Oracle.SqlServer.Postgre.HDFS.Hive.ADS.HBase.TableStore(O ...

  6. 在 C++ 中使用 QML 对象

    看过了如何在 QML 中使用 C++ 类型或对象,现在来看如何在 C++ 中使用 QML 对象. 我们可以使用 QML 对象的信号.槽,访问它们的属性,都没有问题,因为很多 QML 对象对应的类型,原 ...

  7. Prometheus 监控目标运行状态并邮件通知

    Prometheus 监控目标运行状态并邮件通知 邮件服务安装:https://www.cnblogs.com/xiangsikai/p/9809654.html 告警规则示例:https://pro ...

  8. 后台数据转换成Excel,前台下载

    <dependency> <groupId>org.apache.poi</groupId> <artifactId>poi</artifactI ...

  9. C#安全类型转换基于convert

    using Newtonsoft.Json; using System; using System.Collections.Generic; using System.Globalization; n ...

  10. Java 实现生产者 – 消费者模型

    转自:http://www.importnew.com/27063.html 考查Java的并发编程时,手写“生产者-消费者模型”是一个经典问题.有如下几个考点: 对Java并发模型的理解 对Java ...