Oulipo

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 36903		Accepted: 14898

Description

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

3BAPCBAPCAZAAZAZAZAVERDIAVERDXIVYERDIAN

Sample Output

130

Source

/*题意：给你两个字符串P,T让你判断P在T中出线了多少次*//*next数组就是表示以第i位为结尾的字符串，前缀后缀中最长相等字符串的长度*/#include
     
      #include
      
       #include
       
        #define M 1000010using namespace std;int next[M],sum,cur=0;void makeNext(const char P[],int next[]){    int q,k;//q:模版字符串下标；k:最大前后缀长度    int m = strlen(P);//模版字符串长度    next[0]=0;//模版字符串的第一个字符的最大前后缀长度为0    for (q=1,k=0;q
        
         0&&P[q]!=P[k])//递归的求出P[0]···P[q]的最大的相同的前后缀长度k        {            k=next[k-1];          //不理解没关系看下面的分析，这个while循环是整段代码的精髓所在，确实不好理解        }            if (P[q]==P[k])//如果相等，那么最大相同前后缀长度加1        /*仔细想想如果这最后一个字母和遍历到的这个字母都一样了，那么最长的匹配块肯定就会多一个*/        {            k++;        }        next[q]=k;    }}int kmp(const char T[],const char P[],int next[]){    int n,m;    int i,q;    n=strlen(T);    m=strlen(P);    makeNext(P,next);    for(i=0,q=0;i
         
          0&&P[q]!=T[i])            q=next[q-1];        /*这里的while就是参照next数组中的q位置的值，看看需要向前移动多少个单位*/        if (P[q]==T[i])        {            q++;        }        if(q==m)//找到一个完整的字符串        {            cur++;        }    }    }int main(){    freopen("C:\\Users\\acer\\Desktop\\in.txt","r",stdin);    char p[10005],t[1000005];    int n;    scanf("%d",&n);    while(n--)    {        scanf("%s\n%s\n",&p,&t);        cur=0;        memset(next,0,sizeof next);        makeNext(p,next);        kmp(t,p,next);        printf("%d\n",cur);    }    return 0;}