**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题目大意：就是给你一个子串P和一个主串S，求在主串中有多少个子串。。。。
解题思路：这几天一直在整AC自动机，刚开始一看条件反射我以为是AC自动机，结果一想不是，因为，AC自动机都是给你很多个串，让你找前缀的，这个不是，这个是两两比较的所以很明显是KMP，结果就行了。。。。但是刚开始的时候犯了一个错误，后面会给出介绍哦。。。
上代码：
这是AC代码：
#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int maxn = 1000005;
char S[maxn], P[maxn];
int next[maxn];
int ans ;
void Getnext()
{
int j = 0;
int k = -1;
next[0] = -1;
int Plen = strlen(P);
while(j < Plen)
{
if(k==-1 || P[j]==P[k])
{
k++;
j++;
next[j] = k;
}
else
k = next[k];
}
}
int KMP()
{
int i = 0;
int j = 0;
Getnext();
int Slen = strlen(S);
int Plen = strlen(P);
while(i<Slen && j<Plen)
{
if(j==-1 || S[i]==P[j])
{
i++;
j++;
}
else
j = next[j];
if(j == Plen)
{
ans++;
j = next[j];
}
}
return ans;
}
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
ans = 0;
scanf("%s%s",P,S);
KMP();
printf("%d\n",ans);
}
return 0;
}
下面给出一个TLE的代码，是不是感觉与前面几乎一样,请仔细找Bug，其实这也是一种经验啊，说多了都是泪啊。。。。。。；
#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int maxn = 1000005;
char S[maxn], P[maxn];
int next[maxn];
int ans ;
void Getnext()
{
int j = 0;
int k = -1;
next[0] = -1;
while(j < strlen(P))
{
if(k==-1 || P[j]==P[k])
{
k++;
j++;
next[j] = k;
}
else
k = next[k];
}
return ;
}
int KMP()
{
int i = 0;
int j = 0;
Getnext();
while(i<strlen(S) && j<strlen(P))
{
if(j==-1 || S[i]==P[j])
{
i++;
j++;
}
else
j = next[j];
if(j == strlen(P))
{
ans++;
j = next[j];
}
}
return ans;
}
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
ans = 0;
scanf("%s%s",P,S);
KMP();
printf("%d\n",ans);
}
return 0;
}与前面不一样的就是在算字符串的长度的时候，如果一直在while循环里的话，就会一直算，就会TLE的，所以就直接在外面算了，下次一定要注意，应该没有下一次了。。。
