Problem Description Ignatius is building an Online Judge, now he has worked out all the problems except the Judge System. The system has to read data from correct output file and user's result file, then the system compare the two files. If the two f…

题目链接 Problem Description Ignatius is building an Online Judge, now he has worked out all the problems except the Judge System. The system has to read data from correct output file and user's result file, then the system compare the two files. If the…

Online Judge Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 4977    Accepted Submission(s): 1889 Problem Description Ignatius is building an Online Judge, now he has worked out all the problem…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1073 Problem Description Ignatius is building an Online Judge, now he has worked out all the problems except the Judge System. The system has to read data from correct output file and user's result file,…

模拟评测机判断答案 先判断有没有不一样的 有的话再提取出 有效子列 看看有没有错的 #include <iostream> #include <cstdio> #include <cstring> #include <string> using namespace std; int t; string a,b,sa,sb; ]; ][]={"Accepted","Presentation Error","Wr…

分析:水题. #include<iostream> using namespace std; #define N 5050 char a[N],b[N],tmp[N]; void Read(char p[]) { getchar(); gets(tmp); while(gets(tmp)) { if(strcmp(tmp,"END")==0) break; if(strlen(tmp)!=0) strcat(p,tmp); strcat(p,"\n");…

题意:给一系列的输出和标准答案,比较二者是AC,PE或WA 字符串处理还是比较薄弱,目前没什么时间搞字符串专题,所以遇到一题就努力搞懂 #include<cstdio> #include<iostream> #include<algorithm> #include<cstring> #include<cmath> #include<queue> #include<map> using namespace std; #def…

http://acm.hdu.edu.cn/showproblem.php?pid=1073 模拟oj判题 随便搞,开始字符串读入的细节地方没处理好,wa了好久 #include <iostream> #include <cstdio> #include <cstring> #include <map> using namespace std ; ],s2[] ; ],s4[] ; int main() { int t ; scanf("%d%*c…

HDU Today Time Limit: 15000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 26104    Accepted Submission(s): 6341 Problem Description 经过锦囊相助,海东集团终于度过了危机,从此,HDU的发展就一直顺风顺水,到了2050年,集团已经相当规模了,据说进入了钱江肉丝经济开发区500强.这时候,…

Text Reverse                                                                                  Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Problem Description Ignatius likes to write words in reverse way. Given…

Problem Description Ignatius likes to write words in reverse way. Given a single line of text which is written by Ignatius, you should reverse all the words and then output them. Input The input contains several test cases. The first line of the inpu…

2015沈阳区域赛现场赛第2题 题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5510 题意:给定一个由字符串组成的序列,一共n个元素,每个元素是一个不超过2000个字符的字符串.求"存在秩小于 i 且不是 i 的子串"的最大的 i (1<= i <= n). 数据范围:n [1, 500],T组输入,T [1, 50] 思路:从1到n-1枚举每个字符串str[i],判断是否有 j < i 使得str[j]不是str[i…

1073 Scientific Notation(20 分) Scientific notation is the way that scientists easily handle very large numbers or very small numbers. The notation matches the regular expression [+-][1-9].[0-9]+E[+-][0-9]+ which means that the integer portion has exa…

 Hangman Judge  In ``Hangman Judge,'' you are to write a program that judges a series of Hangman games. For each game, the answer to the puzzle is given as well as the guesses. Rules are the same as the classic game of hangman, and are given as follo…

题目大意: 给定一个长度<2000的串,再给最多可达10000的询问区间,求解区间字符串中的不同子串的个数 这里先考虑求解一整个字符串的所有不同子串的方法 对于后缀自动机来说,我们动态往里添加一个字符,每次添加一个字符进去,我们只考虑那个生成的长度为当前长度的后缀自动机的节点 那么这个节点可接收的字符串的个数就是( p->l - p->f->l ),也就是以当前点为最后节点所能得到的与之前不重复的子串的个数 那么这个问题就很好解决了,共2000个位置,以每一个位置为起点构建一次后缀…

不要62 Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 34525    Accepted Submission(s): 12493 Problem Description 杭州人称那些傻乎乎粘嗒嗒的人为62(音:laoer).杭州交通管理局经常会扩充一些的士车牌照,新近出来一个好消息,以后上牌照,不再含有不吉利的数字了,这样一来,就可…

中文意思不解释. 很水,我本来想用switch处理字符串,然后编译不通过...原来switch只能处理整数型的啊,我都忘了. 然后就有了很挫的一大串if代码了... 代码: #include <iostream> #include <string> using namespace std; int digit(string str) { if (str == "zero") return 0; if (str == "one") return…

Problem Description Last year summer Max traveled to California for his vacation. He had a great time there: took many photos, visited famous universities, enjoyed beautiful beaches and tasted various delicious foods. It is such a good trip that Max…

http://acm.hdu.edu.cn/showproblem.php?pid=5083 机器码和操作互相转化 注意SET还要判断末5位不为0输出Error #pragma comment(linker, "/STACK:36777216") #pragma GCC optimize ("O2") #include <cstdio> #include <cstdlib> #include <cmath> #include &l…

题目链接 Problem Description Today is the birthday of SF,so VS gives two strings S1,S2 to SF as a present,which have a big secret.SF is interested in this secret and ask VS how to get it.There are the things that VS tell: Suffix(S2,i) = S2[i...len].Ni is…

题目链接 Problem Description Bob has a dictionary with N words in it. Now there is a list of words in which the middle part of the word has continuous letters disappeared. The middle part does not include the first and last character. We only know the pr…

#include<iostream> #include<stdio.h> #include<string.h> #include<cmath> using namespace std; int main() { int t; char a[55],b[55],r[120];//b插入a中央 scanf("%d",&t); while(t--) { scanf("%s%s",a,b);//可以用gets int…

Parity Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1855    Accepted Submission(s): 1447 Problem Description A bit string has odd parity if the number of 1's is odd. A bit string has even pa…

Online Judge Problem Description Ignatius is building an Online Judge, now he has worked out all the problems except the Judge System. The system has to read data from correct output file and user's result file, then the system compare the two files.…

题目链接 题意 给定两个串\(S,T\),找出\(S\)中所有与\(T\)匹配的子串. 这里,\(T\)的每位上可以有若干(\(\leq 10\))种选择,匹配的含义是:对于\(S\)的子串的每一位,\(T\)的相应位都有一种选择与之对应. 题解 shift-and算法详解 https://www.douban.com/note/321872072/ 搜出来的题解全都是\(shift-and\)的.... 学习了一波...然而并不明白\(kmp\)为什么不可以.... 看到这道题直觉就是和hdu…

How many Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1852    Accepted Submission(s): 763 Problem Description Give you n ( n < 10000) necklaces ,the length of necklace will not large than 10…

题目大意: 不断修改字符串中的字母,然后询问区间字符串是否处于已给定的字符串集合中 这里将原来的字符串集合保存到hash表中,当然用map,set都没有问题 修改查询都用线段树实现,自己的query函数写的有问题,按照网上的改了就没问题 写一下自己的理解,因为左右子树合并的时候,需要计算右子树生成的字符串的长度后,加上base的长度次方 而我们计算右子树中含有的字母数量,靠t-m , 而这个t 必然要处于 l , r 中,向下递归的时候注意修改s,t的区间,不然计算的长度会偏长 #include…

String painter Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2068    Accepted Submission(s): 908 Problem Description There are two strings A and B with equal length. Both strings are made up o…

题目大意: 根据所给的数字,表示其相连的字符的输出个数,或是下一个括号中的所有字符的输出个数 每一个相互对应的 '(' 和 ')' 中的所有字母均作为一组数据处理 在每一次dfs过程中都处理好这样一个对应组中间的字符 再根据前面所带的数字k 循环k次输出自己要的数据 每次输出一个字母,因为无论多复杂的字符串都能够转化成最后只带一个数字和一个字母连接的形式,第38行代码有较详细解释 #include <cstring> #include <cstdio> using namespac…

题面: 魔咒词典 Time Limit: 8000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 14997 Accepted Submission(s): 3588 Problem Description 哈利波特在魔法学校的必修课之一就是学习魔咒.据说魔法世界有100000种不同的魔咒,哈利很难全部记住,但是为了对抗强敌,他必须在危急时刻能够调用任何一个需要的魔咒,所以…