Encrypting passwords is one of the most important problems nowadays, and you trust only the encryp-
tion algorithms which you invented, and you have just made a new encryption algorithm.
Given a password which consists of only lower case English letters, your algorithm encrypts this
password using the following 3 steps (in this given order):
1.
Swap two different characters of the given password (you can do this step zero or more times).
2.
Append zero or more lower case English letters at the beginning of the output of step one.
3.
Append zero or more lower case English letters to the end of the output of step two.
And the encrypted password is the output of step three.
You have just nished implementing the above algorithm and applied it on many passwords. Now
you want to make sure that there are no bugs in your implementation, so you decided to write another
program which validates the output of the encryption program. Given the encrypted password and the
original password, your job is to check whether the encrypted password may be the result of applying
your algorithm on the original password or not.
Input
Your program will be tested on one or more test cases. The rst line of the input will be a single
integer
T
, the number of test cases (1
T
100). Followed by the test cases, each test case is on two
lines. The rst line of each test case contains the encrypted password. The second line of each test case
contains the original password. Both the encrypted password and the original password are at least 1
and at most 100,000 lower case English letters (from `
a
' to `
z
'), and the length of the original password
is less than or equal the length of the encrypted password.
Output
For each test case, print on a single line one word, `
YES
' (without the quotes) if applying the algorithm on
the original password may generate the encrypted password, otherwise print `
NO
' (without the quotes).
SampleOutput
3
abcdef
ecd
cde
ecd
abcdef
fed
SampleOutput
YES
YES
NO//有坑
 
 
 
#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
using namespace std;
const int maxn=;
int hash[maxn];
int cur[maxn];
char s[],t[];
int h[];
int main()
{
int T;
scanf("%d",&T);
getchar();
while(T--)
{
memset(h,,sizeof(h));
memset(hash,,sizeof(hash));
memset(cur,,sizeof(cur));
memset(t,,sizeof(t));
memset(s,,sizeof(s));
scanf("%s",t);
getchar();
scanf("%s",s);
getchar();
int len1=strlen(s);
int len2=strlen(t); for(int i=; i<len1; i++)
{
int temp=s[i];
hash[temp]++;
} int ans=;
int cnt=;
bool flag=false;
if(len1==len2)
{
for(int i=; i<len2; i++)
{
int d=t[i];
h[i]=d;
if(cur[d]<hash[d])
cnt++;
cur[d]++;
if(cnt==len1){
flag=true;
break;
}
}
} else
{
for(int i=; i<len2; i++)
{ int d=t[i];
h[i]=d;
if(cur[d]<hash[d])
cnt++;
cur[d]++; if(i>=len1)
{
d=h[i-len1];
if(cur[d]<=hash[d])
cnt--;
cur[d]--; }
if(cnt==len1){
flag=true;
break;
} }
}
if(flag)
printf("YES\n");
else
printf("NO\n");
}
return ;
}
 
 
 
 
 

哈希UVALive 6326 Contest Hall Preparation的更多相关文章

  1. UVaLive 2796 Concert Hall Scheduling (最小费用流)

    题意:个著名的音乐厅因为财务状况恶化快要破产,你临危受命,试图通过管理的手段来拯救它,方法之一就是优化演出安排,既聪明的决定接受或拒绝哪些乐团的演出申请,使得音乐厅的收益最大化.该音乐厅有两个完全相同 ...

  2. Gym 100952E&&2015 HIAST Collegiate Programming Contest E. Arrange Teams【DFS+剪枝】

    E. Arrange Teams time limit per test:2 seconds memory limit per test:64 megabytes input:standard inp ...

  3. HUNNU--湖师大--11407--It Is Cold

    [F] It Is Cold Dr. Ziad Najem is known as the godfather of  the  ACPC. When the regional contest was ...

  4. 【取对数】【哈希】Petrozavodsk Winter Training Camp 2018 Day 1: Jagiellonian U Contest, Tuesday, January 30, 2018 Problem J. Bobby Tables

    题意:给你一个大整数X的素因子分解形式,每个因子不超过m.问你能否找到两个数n,k,k<=n<=m,使得C(n,k)=X. 不妨取对数,把乘法转换成加法.枚举n,然后去找最大的k(< ...

  5. 【字符串哈希】The 16th UESTC Programming Contest Preliminary F - Zero One Problem

    题意:给你一个零一矩阵,q次询问,每次给你两个长宽相同的子矩阵,问你它们是恰好有一位不同,还是完全相同,还是有多于一位不同. 对每行分别哈希,先一行一行地尝试匹配,如果恰好发现有一行无法对应,再对那一 ...

  6. UVALive - 6893 The Big Painting 字符串哈希

    题目链接: http://acm.hust.edu.cn/vjudge/problem/129730 The Big Painting Time Limit: 5000MS 题意 给你一个模板串和待匹 ...

  7. UVALive - 4671 K-neighbor substrings (FFT+哈希)

    题意:海明距离的定义:两个相同长度的字符串中不同的字符数.现给出母串A和模式串B,求A中有多少与B海明距离<=k的不同子串 分析:将字符a视作1,b视作0.则A与B中都是a的位置乘积是1.现将B ...

  8. HDU 6326.Problem H. Monster Hunter-贪心(优先队列)+流水线排序+路径压缩、节点合并(并查集) (2018 Multi-University Training Contest 3 1008)

    6326.Problem H. Monster Hunter 题意就是打怪兽,给定一棵 n 个点的树,除 1 外每个点有一只怪兽,打败它需要先消耗 ai点 HP,再恢复 bi点 HP.求从 1 号点出 ...

  9. 2016 Asia Jakarta Regional Contest L - Tale of a Happy Man UVALive - 7722

    UVALive - 7722 一定要自己做出来!

随机推荐

  1. 自学youku_web

    仿youku架构 数据库设计 管理员 注册 登录 上传视频 删除视频 发布公告 普通用户 注册 登录 充会员 查看视频 下载免费视频 下载收费视频 查看观影记录 查看公告 思路 class Field ...

  2. 【赛时总结】 ◇赛时·III◇ AtCoder ABC-099

    ◆赛时·III◆ ABC-099 ■唠叨■ 不要问我为什么先给ABC-100写了博客再写的ABC-099-- 莫名觉得这次比赛特别简单--虽然我并没有参加比赛,只是之后再补做的.QwQ ■试题& ...

  3. SpringMVC-实现PUT请求上传文件(转)

    因为在图片上传的时候使用的是二进制的方式上传,所以使用隐藏域进行方法转换方式失效,转方法: https://www.cnblogs.com/morethink/p/6378015.html 可是后来我 ...

  4. scrapy--json(喜马拉雅Fm)(二)

    学习了对数据的储存,感觉还不够深入,昨天开始对储存数据进行提取.整合和图像化显示.实例还是喜马拉雅Fm,算是对之前数据爬取之后的补充. 明确需要解决的问题 1,蕊希电台全部作品的进行储存 --scra ...

  5. python--Matplotlib(二)

    Matplotlib+pandas作图 一.对csv文件进行提取ruixi.csv 对上述表格进行提取并做图 #-*- coding:utf-8 -*- import matplotlib as mp ...

  6. myql简单语法测试

    删除某一行  delete from name1 where agee=10 limit 1; insert into name1(agee,namee)values(10,'wwww'),(10,' ...

  7. 【linux】【指令】systemctl 指令部分解读

    systemctl [OPTIONS...] {COMMAND} ... Query or send control commands to the systemd manager. -h --hel ...

  8. B1018 锤子剪刀布 (20分)

    B1018 锤子剪刀布 (20分) 大家应该都会玩"锤子剪刀布"的游戏:两人同时给出手势. 现给出两人的交锋记录,请统计双方的胜.平.负次数,并且给出双方分别出什么手势的胜算最大. ...

  9. python——获取数据类型

    在python中,可使用type()和isinstance()内置函数获取数据类型 如: (1)type()的使用方法: >>> a = '230'         >> ...

  10. POJ-3126 BFS,埃式筛选及黑科技

    题目大意:给定两个四位素数a  b,要求把a变换到b,变换的过程要保证  每次变换出来的数都是一个 四位素数,而且当前这步的变换所得的素数  与  前一步得到的素数  只能有一个位不同,而且每步得到的 ...