B - Oulipo

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. 
 

Input

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.

Output

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. 
 

Sample Input

3
BAPC
BAPC
AZA
AZAZAZA
VERDI
AVERDXIVYERDIAN

Sample Output

1
3
0

#include<bits/stdc++.h>

using namespace std;

const int maxn=1e4+10;
const int N=1e6+4;
char pattern[maxn];
char text[N];
int prefix[maxn];
int n;
int m;
int ans;
void prefix_table()//记录字符，从下标0开始记录前缀和，，字符的长度 
{
	prefix[0]=0;

	int len=0;
	int i=1;
	while(i<n){ 
		if(pattern[len]==pattern[i]){
			len++;
			prefix[i]=len;
			i++;
		}
		else{
			if(len>0){
				len=prefix[len-1];
			}
			else{
				prefix[i]=len;
				i++;
			}
		}
	}
}
void move_prefix_table(){
	for(int i=n-1;i>=1;i--){
		prefix[i]=prefix[i-1];
	}
	prefix[0]=-1;
}
void kmp_search(){
	prefix_table();
	move_prefix_table();
	int i=0;//text的下标 
	int j=0;//pattern的下标 
	while(i<m){
		if(j==n-1&&pattern[j]==text[i]){
			ans++;//记录子串的个数 
			j=prefix[j];
		}
		if(pattern[j]==text[i]){
			j++;i++;
		}
		else{
			j=prefix[j];
			if(j==-1){
				j++;i++;
			}
		}
	}
}
int main()
{
	int t;
	scanf("%d",&t);
	while(t--){
		ans=0;
		memset(prefix,0,sizeof(prefix));
		scanf("%s%s",pattern,text);
		n=strlen(pattern);
		m=strlen(text);
		kmp_search();
		printf("%d\n",ans); 
	}
	return 0;
}

用kmp还是超时了，不知道为什么还望有大神能指出来，非常感谢