Oulipo

he 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

3BAPCBAPCAZAAZAZAZAVERDIAVERDXIVYERDIAN

Sample Output

13
0
题目意思：给你一个子串，求该子串在主串中出现的次数。
解题思路：利用kmp，不断匹配，匹配到了计数器就加1，但是这个题我第一次写超时了，因为一次匹配成
           功后，我就将n置为0，让主串倒退到上一个出现子串开头的地方。
           后来改进了一下，主串其实不用倒退，只需将子串倒退到next[n]的位置，即可！
#include<iostream>#include<cstring>using namespace std;int const maxn=1000005;char str[maxn],str1[maxn];int next1[maxn];int len,len1;void sign(){int i,j;next1[0]=-1;i=-1,j=0;int len1=strlen(str1);while(j<len1){    if(i==-1||str1[i]==str1[j])    {    i++;    j++;    next1[j]=i;}else{i=next1[i];}}}int kmp(){int m=0,n=0,num=0;int len=strlen(str);int len1=strlen(str1);sign();while(m<len){if(n==-1||str[m]==str1[n]){    m++;n++;}else{n=next1[n];}if(n==len1){n=next1[n];num++;}}return num;}int main(){int T;cin>>T;while(T--){scanf("%s",str1);   //子串scanf("%s",str);    //主串int num1=kmp();cout<<num1<<endl; }return 0;}