POJ 3461 Oulipo

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.

---

【题目分析】
出现了多少次，用KMP算法预处理一下模板，然后再匹配一遍。当匹配长度大于匹配串的长度时。就给答案加上1。最后输出就好了。

---

【代码】

#include <cstdio>#include <cstring>char s[1000002],ss[10002];int next[1000002];int main(){    int tt;    scanf("%d",&tt);    while (tt--)    {        scanf("%s",ss+1);        scanf("%s",s+1);        int lens=strlen(s+1),lenss=strlen(ss+1);        next[1]=0;int j=0;        for (int i=2;i<=lens;++i)        {            while (j>0&&s[j+1]!=s[i]) j=next[j];            if (s[j+1]==s[i]) j++;            next[i]=j;        }        int ans=0; j=0;        for (int i=1;i<=lens;++i)        {            while (j>0&&ss[j+1]!=s[i]) j=next[j];            if (ss[j+1]==s[i]) j++;            if (j==lenss) ans++,j=next[j];        }        printf("%d\n",ans);    }}