hdu 1686 Oulipo(简单KMP)只不过比赛的时候用了string.一直超时，改成char就一遍AC，纠结。。。

1、http://acm.hdu.edu.cn/showproblem.php?pid=1686

2、题目大意：

这道题目太纠结了，就是一道直接用KMP模板的题目，比赛的时候居然第一次想成做for循环次的KMP，超时了，后来知道只能用一次KMP了，结果用的string 一直超时，后来改成char居然一遍AC，还得好好熟悉算法

题目是给出两个字符串a,b,求a在b中出现多少次

3、题目：

Oulipo

Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 3999 Accepted Submission(s): 1578



[align=left]Problem Description[/align]
The French author Georges Perec (1936–1982) once wrote a book, La disparition, without the letter 'e'. He was a member of the Oulipo group. A quote from the book:

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.

[align=left]Input[/align]
The 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.

[align=left]Output[/align]
For 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.

[align=left]Sample Input[/align]

3
BAPC
BAPC
AZA
AZAZAZA
VERDI
AVERDXIVYERDIAN

[align=left]Sample Output[/align]

1
3
0

[align=left]Source[/align]
华东区大学生程序设计邀请赛_热身赛

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4、ＡＣ代码：

#include<stdio.h>
#include<iostream>
#include<string.h>
#include<string>
#include<algorithm>
using namespace std;
char s[1000005];
char a[10005];
char b[1000005];
int f[10005];
void getFail(char p[], int *f)
{
int m=strlen(p);
f[0]=f[1]=0;
for(int i=1; i<m; ++i)
{
int j=f[i];
while(j && p[i]!=p[j])j=f[j];
f[i+1] = p[i]==p[j]?1+j:0;
}
}
int find(char T[],char p[],int *f)
{
getFail(p,f);
int n=strlen(T);
int m=strlen(p);
int j=0;
int sum=0;
for(int i=0; i<n; ++i)
{
while(j && T[i]!=p[j])j=f[j];
if(T[i]==p[j])++j;
if(j==m)
{
sum++;
}
}
return sum;
}
int main()
{
int n;
scanf("%d",&n);
while(n--)
{
scanf("%s",a);
scanf("%s",b);
int la=strlen(a);
int lb=strlen(b);
int sum=find(b,a,f);
printf("%d\n",sum);
}
return 0;
}

附string 超时代码：

#include<iostream>
#include<string.h>
#include<string>
#include<algorithm>
using namespace std;
char s[1000005];
string a;
string b;
int f[10005];
void getFail(string p, int *f)
{
int m=p.length();
f[0]=f[1]=0;
for(int i=1; i<m; ++i)
{
int j=f[i];
while(j && p[i]!=p[j])j=f[j];
f[i+1] = p[i]==p[j]?1+j:0;
}
}
int find(string T,string p,int *f)
{
getFail(p,f);
int n=T.length();
int m=p.length();
int j=0;
int sum=0;
for(int i=0; i<n; ++i)
{
while(j && T[i]!=p[j])j=f[j];
if(T[i]==p[j])++j;
if(j==m)
{
sum++;
}
}
return sum;
}
int main()
{
int n;
cin>>n;
while(n--)
{
cin>>a;
cin>>b;
int la=a.length();
int lb=b.length();
int sum=find(b,a,f);
//printf("%d\n",sum);
cout<<sum<<endl;
}
return 0;
}