Easier Done Than Said?                                                                     Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Problem Description Password security is a tricky thing. Users prefer simpl…

Problem Description Password security is a tricky thing. Users prefer simple passwords that are easy to remember (like buddy), but such passwords are often insecure. Some sites use random computer-generated passwords (like xvtpzyo), but users have a…

Easier Done Than Said? Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 16543    Accepted Submission(s): 7846 Problem Description Password security is a tricky thing. Users prefer simple password…

Easier Done Than Said? Problem Description Password security is a tricky thing. Users prefer simple passwords that are easy to remember (like buddy), but such passwords are often insecure. Some sites use random computer-generated passwords (like xvtp…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1039 Problem Description Password security is a tricky thing. Users prefer simple passwords that are easy to remember (like buddy), but such passwords are often insecure. Some sites use random computer-g…

水水的 #include <iostream> #include <cstring> using namespace std; ]; bool flag; int vol,v2,con; int main() { while(cin>>s) { if(!strcmp(s,"end")) break; flag=; v2=vol=con=; ;s[i]!='\0';i++) { ]&&(s[i]!='e'&&s[i]!=…

规则: 1.必须至少包含一个元音字母.a e i o u 2.不能包含三个连续元音或者连续辅音字母. 3.不能包含两个连续字母,除了'ee'和'oo'. PS:字母个数(1<= N <=20). #define _CRT_SECURE_NO_DEPRECATE #include <stdio.h> #include <stdlib.h> #include <string.h> int is_vowel(char strIn) { if(strIn == 'a…

题意是检查一个字符串是否满足三个条件: 一.至少有一个元音字母.二.不能出现三个连续的元音或三个连续的辅音.三.除了 ee 和 oo 外不能出现两个连续相同字母. 若三个条件都能满足,该字符串满足条件,有一个条件不满足则该字符串不满足条件. 但是这道题的数据......一定有元音字母,长度一定不少于 3.省去很多麻烦...... 代码如下: #include <bits/stdc++.h> using namespace std; int main() { int len; bool f; s…

题目链接:hdu 5469 Antonidas 题意: 给你一颗树,每个节点有一个字符,现在给你一个字符串S,问你是否能在树上找到两个节点u,v,使得u到v的最短路径构成的字符串恰好为S. 题解: 这题可以用树的分治+字符串hash,不过搜索+剪枝写的好一样可以过,而且跑的时间和正解差不多. 搜索的做法就是先随便找一个点当作根,然后预处理一下最大的深度,然后枚举起点,开始向各个方向遍历,如果这个点的最大深度小于未匹配的字符串长度,那么久向父亲方面搜. #include<bits/stdc++.h…

http://acm.hdu.edu.cn/showproblem.php?pid=2087 Problem Description 一块花布条,里面有些图案,另有一块直接可用的小饰条,里面也有一些图案.对于给定的花布条和小饰条,计算一下能从花布条中尽可能剪出几块小饰条来呢?   Input 输入中含有一些数据,分别是成对出现的花布条和小饰条,其布条都是用可见ASCII字符表示的,可见的ASCII字符有多少个,布条的花纹也有多少种花样.花纹条和小饰条不会超过1000个字符长.如果遇见#字符,则不…