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.

 

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.

 

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.

 

#include<cstdio>#include<cstring>#include<iostream>using namespace std;int n,len1,len2,f[1000001],ans;char s1[1000001],s2[1000001];void getnext(){    for(int i=1;i<len1;i++)    {        int j=f[i];        while(j&&s1[i]!=s1[j]) j=f[j];        f[i+1]= s1[i]==s1[j] ? j+1 : 0;    }}void getans(){    int j=0;    for(int i=0;i<len2;i++)    {        while(j&&s1[j]!=s2[i]) j=f[j];        if(s1[j]==s2[i]) j++;        if(j==len1) ans++;    }    printf("%d\n",ans);}int main(){    scanf("%d",&n);    while(n--)    {        cin>>s1>>s2;        len1=strlen(s1);len2=strlen(s2);        memset(f,0,sizeof(f));        getnext();        getans();        ans=0;    }}

//由于字符串位置从0开始存//所以f[i]在数值上=最大前缀子串=i的最大后缀子串=下一个要检验的位置 //也就是保证了 0——f[i]-1是相等的，再匹配i与f[i] #include<cstdio>#include<cstring>#include<iostream>using namespace std;int n,len1,len2,f[1000001],ans;char s1[1000001],s2[1000001];void getnext(){    for(int i=1;i<len1;i++)//注意这里从1开始     {        int j=f[i];//j=下一个要匹配的位置         while(j&&s1[i]!=s1[j]) j=f[j];        f[i+1]= s1[i]==s1[j] ? j+1 : 0;//j+1：字符串从0开始，所以长度+1     }}void getans(){    int j=0;    for(int i=0;i<len2;i++)//这里从1开始     {        while(j&&s1[j]!=s2[i]) j=f[j];        if(s1[j]==s2[i]) j++;        if(j==len1) ans++;    }    printf("%d\n",ans);}int main(){    scanf("%d",&n);    while(n--)    {        cin>>s1>>s2;        len1=strlen(s1);len2=strlen(s2);        memset(f,0,sizeof(f));        getnext();        getans();        ans=0;    }}