1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 using namespace std; 5  6 const int N = 1000001; 7 int n,m,nxt[N],j; 8 char s1[N],s2[N]; 9 10 void INT() {11     nxt[0]=0,nxt[1]=0;12     for(int i=1; i<m; i++) { 13         j = nxt[i];14         while(j>0 && s2[i]!=s2[j]) j=nxt[j];15         if(s2[i]==s2[j]) nxt[i+1]=j+1;16         else nxt[i+1] = 0;17     } 18 }19 20 void find() {21     j=0;22     for(int i=0; i<n; i++) {23         while(j>0&&s1[i]!=s2[j])  j=nxt[j];24         if(s1[i]==s2[j]) j++;25         if(j == m) 26             cout<<i-m+2<<endl;27     }28 }29 30 int main() {31     /*gets(s1);32      gets(s2);*/ //无情的删去33      cin>>s1>>s2;34     n=strlen(s1); m=strlen(s2);35     INT();36     find();37     for(int i=1; i<=m; i++)38         cout<<nxt[i]<<" ";39     return 0;40 }

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.

 1 #include<cstdio> 2 #include<cstring> 3 #include<iostream> 4 using namespace std; 5 char t[1000001],p[1000001]; 6 int lt,lp,f[1000001],ans,s; 7 void INT() 8 { 9     f[1]=0;10     for(int i=1;i<lp;i++)11     {12         int j=f[i];13         while(j&&p[i]!=p[j]) j=f[j];14         f[i+1]= p[i]==p[j] ? j+1 : 0;15     }16 }17 void Find()18 {19     int j=0;20     for(int i=0;i<lt;i++)21     {22         while(j&&(p[j]!=t[i])) j=f[j];23         if(p[j]==t[i]) j++;24         if(j==lp) ans++;25     }26 }27 int main()28 {29     cin>>s;30     while(s--) {31         ans=0;32         memset(f,0,sizeof(f));33         cin>>p>>t;34         lt=strlen(t);35         lp=strlen(p);36         INT();37         Find();38         cout<<ans<<endl;39     }40     return 0;41 }