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。。。

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.

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.

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.

  
3
BAPC
BAPC
AZA
AZAZAZA
VERDI
AVERDXIVYERDIAN

  
1
3
0

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;
}

#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;
}