What Are You Talking About Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 102400/204800 K (Java/Others) Total Submission(s): 16042    Accepted Submission(s): 5198 Problem Description Ignatius is so lucky that he met a Martian yesterday. But…

What Are You Talking About Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 102400/204800 K (Java/Others) Total Submission(s): 20680    Accepted Submission(s): 6852 Problem Description Ignatius is so lucky that he met a Martian yesterday. But…

What Are You Talking About Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 102400/204800 K (Java/Others) Total Submission(s): 15966    Accepted Submission(s): 5177 Problem Description Ignatius is so lucky that he met a Martian yesterday. But…

题意:单词翻译 思路:map #include<iostream> #include<stdio.h> #include<string.h> #include<map> using namespace std; map<string,string>mp; int main(){ string s1,s2; char c; cin>>s1; while(cin>>s1){ if(s1=="END")bre…

[python]Leetcode每日一题-前缀树(Trie) [题目描述] Trie(发音类似 "try")或者说 前缀树 是一种树形数据结构,用于高效地存储和检索字符串数据集中的键.这一数据结构有相当多的应用情景,例如自动补完和拼写检查. 请你实现 Trie 类: Trie() 初始化前缀树对象. void insert(String word) 向前缀树中插入字符串 word . boolean search(String word) 如果字符串 word 在前缀树中,返回 tru…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1052 Problem Description Here is a famous story in Chinese history. "That was about 2300 years ago. General Tian Ji was a high official in the country Qi. He likes to play horse racing with the king and…

Super Jumping! Jumping! Jumping! Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 49436    Accepted Submission(s): 22871 Problem Description Nowadays, a kind of chess game called "Super Jumping!…

统计难题Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131070/65535 K (Java/Others)Total Submission(s): 50524    Accepted Submission(s): 17827 Problem DescriptionIgnatius最近遇到一个难题,老师交给他很多单词(只有小写字母组成,不会有重复的单词出现),现在老师要他统计出以某个字符串为前缀的单词数量(单词本身也是自己的前缀…

pid=1671">Phone List Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 12879    Accepted Submission(s): 4391 Problem Description Given a list of phone numbers, determine if it is consistent in…

<题目链接> 题目大意: 意思是,从第1个字母到第2字母组成的字符串可由某一周期性的字串(“a”) 的两次组成,也就是aa有两个a组成: 第三行自然就是aabaab可有两个aab组成: 第四行aabaabaab可由三个aab组成: 第五行aabaabaabaab可有四个aab组成 解题分析: 求字符串的前缀是否为周期串,若是,打印循环节的长度及循环次数: #include <cstdio> #include <cstring> ; char s[M]; int nxt[…

Description John is the only priest in his town. October 26th is the John's busiest day in a year because there is an old legend in the town that the couple who get married on that day will be forever blessed by the God of Love. This year N couples p…

Parentheses Matrix Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 0    Accepted Submission(s): 0Special Judge Problem Description A parentheses matrix is a matrix where every element is eithe…

题意: 给你一个1e9-1e14的质数P,让你找出这个质数的前一个质数Q,然后计算Q!mod P 题解: 1e14的数据范围pass掉一切素数筛法,考虑Miller-Rabin算法. 米勒拉宾算法是一种判断素数的随机化算法,由于其随机性,它不能保证总是正确的,但其对于一个素数,总会返回素数的结果,对于一个合数,才有极小概率返回素数的结果(假阳性). 米勒拉宾算法对于单个素数的判断时间复杂度为$O(log^3n)$.(因为1e14相乘会爆longlong,模乘要写成龟速乘,因此要多一个log) 1…

题意: N=a[1]+a[2]+a[3]+...+a[m];  a[i]>0,1<=m<=N; 例如: 4 = 4;  4 = 3 + 1;  4 = 2 + 2;  4 = 2 + 1 + 1;  4 = 1 + 1 + 1 + 1; 共有5种. 给N,问共有几种构造方式. 思路: 一个数N分解的式子中1的个数可以是0,1,2,3,...,N. 2的个数可以是0,1,2,...,N/2. .... 母函数基础题,, 看代码. 当然也可以用DP(背包) 母函数代码: int N,num;…

题意: 有s个不同的单词,给出一个长字符串把这个字符串分解成若干个单词的连接(可重复使用),有多少种分解方法 分析: dp[i]表示i开始的字符串能分解的方法数 dp[i]=sum(dp[i+len(x)]);单词x是i开始的字符串的前缀. #include <map> #include <set> #include <list> #include <cmath> #include <queue> #include <stack> #…

What Are You Talking About Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 102400/204800 K (Java/Others)Total Submission(s): 12617    Accepted Submission(s): 4031 Problem Description Ignatius is so lucky that he met a Martian yesterday. But…

刚学的字典树,代码写得很不熟练.写法上也没有什么特别的优化,就是以1A为第一目标! 可惜还是失败了. 少考虑了一种情况,就是一个单词是另一个单词前缀的问题,写了好久,还是没有1A.不过感觉对字典树有了更深刻的理解. 代码写得不好,看看别人都是怎么写字典树的去. #include<stdio.h> #include<string.h> #include<malloc.h> struct node { node *a[27]; char s[15]; }; char s1[…

What Are You Talking About Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 102400/204800 K (Java/Others) Total Submission(s): 28482    Accepted Submission(s): 9710 Problem Description Ignatius is so lucky that he met a Martian yesterday. But…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4864 解题报告:有n台机器用来完成m个任务,每个任务有一个难度值和一个需要完成的时间,每台机器有一个可以工作的最长时间和一个可以完成的任务的难度的最大值, 一台机器能完成一个任务的条件是这台机器的最长工作时间和能完成任务的难度值必须都大于等于这个任务,而且一台机器最多完成一个任务,假设一个任务的时间为t,难度值为x,那么完成这个任务可以赚到的钱 money = 500 * t + 2 * x; 现在…

http://acm.hdu.edu.cn/showproblem.php?pid=1540 Tunnel Warfare Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 3168    Accepted Submission(s): 1216 Problem Description During the War of Resistan…

题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=1203 I NEED A OFFER! Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 33303    Accepted Submission(s): 13470 Problem Description Speakless很早就想出国,现在…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4956 Poor Hanamichi Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 7    Accepted Submission(s): 4 Problem Description Hanamichi is taking part in…

How many integers can you find Time Limit:5000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Submit Status Practice HDU 1796 Description   Now you get a number N, and a M-integers set, you should find out how many integers which are sm…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4576 坑大发了,居然加 % 也会超时: #include <cstdio> #include <iostream> #include <sstream> #include <cmath> #include <cstring> #include <cstdlib> #include <string> #include <v…

http://acm.hdu.edu.cn/showproblem.php?pid=2846 题意:给出N个模式串,再给出M个文本串,问每一个文本串在多少个模式串中出现. 思路:平时都是找前缀的,这里将模式串s[1……len]的每一个[i,len]的子串都插入,这样就可以满足条件.还要注意如果两个子串都为同一个模式串的子串,不能重复计数.可以用一个id数组装上一次是哪个串的id,如果id相同就不要重复计数了. #include <cstdio> #include <algorithm&g…

http://acm.hdu.edu.cn/showproblem.php?pid=4825 题意:给出N个数,M个询问,每个询问给出一个X,问在这N个数中哪个数和X异或后结果最大. 思路:可以用Trie构造出sigmaSize为0和1的点,先将N个数插入Trie,然后询问在Trie上尽量找可以不同的数,不同的话异或起来是最大的. #include <cstdio> #include <algorithm> #include <iostream> #include &l…

http://acm.hdu.edu.cn/showproblem.php?pid=1024 Max Sum Plus Plus Problem Description Now I think you have got an AC in Ignatius.L's "Max Sum" problem. To be a brave ACMer, we always challenge ourselves to more difficult problems. Now you are fac…

题目链接 题意 : 从(0,0)点走到(N-1,M-1)点,问最少时间. 思路 : BFS..... #include <stdio.h> #include <string.h> #include <queue> #include <iostream> using namespace std ; struct node { int x,y ; int tim ; friend bool operator < (node a,node b) { retu…

题意: 给你一组数,开始询问给一个数  求组中与该数异或值最大的数. 分析:根据异或的特点 要想得到的异或值最大 尽可能的让两个数的每位都相反 先把给定的一组数建树,数的最后一位对应的节点保存这个数的位置(放便取) 对于每个询问 在搜树时优先考虑和当前数位相反的节点. #include <map> #include <set> #include <list> #include <cmath> #include <queue> #include &…

Problem Description Baby Ming collected lots of cell phone numbers, and he wants to sell them for money. He thinks normal number can be sold for b yuan, while number with following features can be sold for a yuan. .The last five numbers are the same.…