C - Oulipo
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.
InputThe first line of the input file contains a single number: the number of test cases to follow. Each test case has the following format:
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.
OutputFor every test case in the input file, the output should contain a single number, on a single line: the number of occurrences of the word W in the text T.
Sample Input
3
BAPC
BAPC
AZA
AZAZAZA
VERDI
AVERDXIVYERDIAN
Sample Output
1
3
0
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
int Next[1100000];
char s1[1100000],s2[1100000];
void getnext(int l)
{
int i=0,j=-1;
Next[0]=-1;
while(i<l)
{
if(j==-1||s1[i]==s1[j])
{
i++,j++;
Next[i]=j;
}
else
j=Next[j];
}
}
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
scanf("%s%s",s1,s2);
int l1=strlen(s1);
int l2=strlen(s2);
if(l1>l2)
{
printf("0\n");
continue;
}
getnext(l1);
int i=0,j=0,ans=0;
while(i<l2)
{
if(j==-1||s1[j]==s2[i])
{
i++,j++;
}
else j=Next[j];
if(j==l1)
{
j=Next[j];
ans++;
}
}
printf("%d\n",ans);
}
}
C - Oulipo的更多相关文章
- C++之路进阶——poj3461(Oulipo)
Oulipo Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 35694 Accepted: 14424 Descript ...
- poj3461 Oulipo(KMP模板)
Oulipo Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 17795 Accepted: 7160 Descripti ...
- 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 < ...
- KMP算法 hdu4686 Oulipo
Problem Description The French author Georges Perec (1936–1982) once wrote a book, La disparition, w ...
- hdu----1686 Oulipo (ac自动机)
Oulipo Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Subm ...
- 字符串hash - POJ 3461 Oulipo
Oulipo Problem's Link ---------------------------------------------------------------------------- M ...
- POJ 3461 Oulipo
E - Oulipo Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u Submit ...
- Oulipo (kmp)
Oulipo Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 26857 Accepted: 10709 Descript ...
- 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 ...
- A - Oulipo
A - Oulipo Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u Submit S ...
随机推荐
- 【高精度】计算2的N次方
题目相关 [题目描述] 任意给定一个正整数N(N≤100),计算2的n次方的值. [输入] 输入一个正整数N. [输出] 输出2的N次方的值. [输入样例] 5 [输出样例] 32 分析 本题考察的是 ...
- Flutter 布局类组件:流式布局(Wrap和Flow)
前言 把超出屏幕显示范围会自动折行的布局称为流式布局.Flutter中通过Wrap和Flow来支持流式布局,将Row换成Wrap后溢出部分则会自动折行. Wrap 接口描述 Wrap({ Key ke ...
- 解决Tengine健康检查引起的TIME_WAIT堆积问题
简介: 解决Tengine健康检查引起的TIME_WAIT堆积问题 一. 问题背景 "服务上云后,我们的TCP端口基本上都处于TIME_WAIT的状态"."这个问题在线下 ...
- linux之平均负载(学习笔记非原创)
什么是平均负载 [root@111 ~]# uptime 11:03:33 up 149 days, 17:34, 1 user, load average: 0.08, 0.05, 0.01 最后三 ...
- mysql .sock丢时候如何链接数据库
在mysql服务器本机上链接mysql数据库时,经常会噢出现mysql.sock不存在,导致无法链接的问题,这是因为如果指定localhost作为一个主机名,则mysqladmin默认使用unix套接 ...
- 【Jboss】应用中缺少宋体怎么办
环境jboss4.2.2 系统CentOS7.2 1.新搭建的环境,但是没有字符集,在windows上的电脑上复制了一份宋体,打成zip包 将zip包上传到服务器中,解压 2.在/usr/share/ ...
- 使用 gRPCurl 调试.NET 5的gPRC服务
介绍 你用过 Curl 吗?这个工具允许你通过 http 来发送数据,现在有一个适用于gGRPC的工具,gRPCurl,在本文中,我将介绍如何下载安装这个工具,然后通过这个工具调试我们.NET 5上面 ...
- ctfshow—web—web签到题
打开靶机,发现只有一句话 查看源码 发现一段字符串,猜是base64加密 拿到flag
- Array.of使用实例
Array.of是es6新增的API,其实粗暴点理解,光看of,就可以猜到它是数组的意思,所以猜测可以用来把字符串转换成数组. 像这样的table,有批量删除和单个删除的功能,,但是又不想写两个方法, ...
- Django Signals
信号 Django中提供了"信号调度",用于在框架执行操作时解耦.通俗来讲,就是一些动作发生的时候,信号允许特定的发送者去提醒一些接受者. Django内置的信号 Model si ...