(KMP)Oulipo -- poj --3461
http://acm.hust.edu.cn/vjudge/contest/view.action?cid=92486#problem/B
http://poj.org/problem?id=3461
Description
Tout avait Pair normal, mais tout s’affirmait faux. Tout avait Fair normal, d’abord, puis surgissait l’inhumain, l’affolant. Il aurait voulu savoir où s’articulait l’association qui l’unissait au roman : stir son tapis, assaillant à tout instant son imagination, l’intuition d’un tabou, la vision d’un mal obscur, d’un quoi vacant, d’un non-dit : la vision, l’avision d’un oubli commandant tout, où s’abolissait la raison : tout avait l’air normal mais…
Perec would probably have scored high (or rather, low) in the following contest. People are asked to write a perhaps even meaningful text on some subject with as few occurrences of a given “word” as possible. Our task is to provide the jury with a program that counts these occurrences, in order to obtain a ranking of the competitors. These competitors often write very long texts with nonsense meaning; a sequence of 500,000 consecutive 'T's is not unusual. And they never use spaces.
So we want to quickly find out how often a word, i.e., a given string, occurs in a text. More formally: given the alphabet {'A', 'B', 'C', …, 'Z'} and two finite strings over that alphabet, a word W and a text T, count the number of occurrences of W in T. All the consecutive characters of W must exactly match consecutive characters of T. Occurrences may overlap.
Input
One line with the word W, a string over {'A', 'B', 'C', …, 'Z'}, with 1 ≤ |W| ≤ 10,000 (here |W| denotes the length of the string W).
One line with the text T, a string over {'A', 'B', 'C', …, 'Z'}, with |W| ≤ |T| ≤ 1,000,000.
Output
Sample Input
Sample Output
3
0
#include<stdio.h>
#include<string.h>
#include<stdlib.h> #define N 1000007 char M[N], S[N];
int Next[N], sum; void FindNext(int Slen)
{
int i=, j=-;
Next[] = -; while(i<Slen)
{
if(j==- || S[i]==S[j])
Next[++i] = ++j;
else
j = Next[j];
}
} void KMP(int Slen, int Mlen)
{
int i=, j=; FindNext(Slen); while(i<Mlen)
{
while(j==- || (M[i]==S[j] && i<Mlen && j<Slen))
i++, j++;
if(j==Slen)
sum++;
j = Next[j];
}
} int main()
{
int t;
scanf("%d", &t);
while(t--)
{
scanf("%s%s", S, M); int Slen = strlen(S), Mlen = strlen(M); sum = ; KMP(Slen, Mlen); printf("%d\n", sum);
}
return ;
}
(KMP)Oulipo -- poj --3461的更多相关文章
- HDU 1686 Oulipo / POJ 3461 Oulipo / SCU 2652 Oulipo (字符串匹配,KMP)
HDU 1686 Oulipo / POJ 3461 Oulipo / SCU 2652 Oulipo (字符串匹配,KMP) Description The French author George ...
- Oulipo POJ - 3461(kmp,求重叠匹配个数)
Problem Description The French author Georges Perec (1936–1982) once wrote a book, La disparition, w ...
- Match:Oulipo(POJ 3461)
Oulipo 题目大意:给你一个字符串,要你找到字符串包含指定子串的个数 只要你知道了KMP,这一题简直不要太简单,注意STL的string是会超时的,还是乖乖用char吧 #include < ...
- AC日记——Oulipo poj 3461
Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 37958 Accepted: 15282 Description The ...
- POJ 3461 Oulipo[附KMP算法详细流程讲解]
E - Oulipo Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u Submit ...
- POJ 3461 Oulipo
E - Oulipo Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u Submit ...
- POJ 3461 Oulipo(乌力波)
POJ 3461 Oulipo(乌力波) Time Limit: 1000MS Memory Limit: 65536K [Description] [题目描述] The French autho ...
- POJ 3461 Oulipo 【KMP统计子串数】
传送门:http://poj.org/problem?id=3461 Oulipo Time Limit: 1000MS Memory Limit: 65536K Total Submission ...
- POJ 3080 Blue Jeans、POJ 3461 Oulipo——KMP应用
题目:POJ3080 http://poj.org/problem?id=3080 题意:对于输入的文本串,输出最长的公共子串,如果长度相同,输出字典序最小的. 这题数据量很小,用暴力也是16ms,用 ...
随机推荐
- js中常用的内部函数的使用
<!DOCTYPE html><html lang="en"><head> <meta charset="UTF-8" ...
- SQL 数据库 子查询及示例
子查询,又叫做嵌套查询. 将一个查询语句做为一个结果集供其他SQL语句使用,就像使用普通的表一样,被当作结果集的查询语句被称为子查询. 子查询有两种类型: 一种是只返回一个单值的子查询,这时它可以用在 ...
- input:file属性
1.accept 只能选择png和gif图片 <input id="fileId1" type="file" accept="image/png ...
- RedisTemplate Redis 操作
stringRedisTemplate.opsForValue().set("test", "100",60*10,TimeUnit.SECONDS);//向r ...
- Spring Boot 连接池
配置方法 基于当前的1.5.2.RELEASE的Spring Boot. 依照官方文档,如果增加了如下依赖的配置,或者类路径中存在spring-boot-starter-jdbc的jar,那么已默认启 ...
- Nexus 按项目类型分配不同的工厂来发布不同 项目
但是有时候,一个公司会有很多项目[crm,oa,erp]等等的项目.如果把这些项目全部都放到releases或者snapshots中的话会有点混乱.比较好的办法是,按项目来分.每个项目一个工厂:cms ...
- Linux Tomcat重新启动
在Linux系统下,重启Tomcat使用命令操作的! 首先,进入Tomcat下的bin目录 cd /usr/local/tomcat/bin 使用Tomcat关闭命令 ./shutdown.sh 查看 ...
- inner outer
left join 是left outer join的简写,left join默认是outer属性的. Inner JoinInner Join 逻辑运算符返回满足第一个(顶端)输入与第二个(底端)输 ...
- Layouts
[Layouts] Each layout file must contain exactly one root element, which must be a View or ViewGroup ...
- luoguP1064 金明的预算方案 (有依赖的背包问题)
题目链接:https://www.luogu.org/problemnew/show/P1064 这是一个有依赖的背包问题,属于01背包的变式.这题还好,每个主件最多有2个附件,那么在对主件进行背包的 ...