Oulipo HDU - 1686

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.

InputThe 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. 
OutputFor 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

3
BAPC
BAPC
AZA
AZAZAZA
VERDI
AVERDXIVYERDIAN

Sample Output

1
3
0

不会kmp的我mmp 
网上的题解全是kmp  
我觉得hash暴力也应该也行

实践证明 暴力hash是可以的

 #include <stdio.h>
 #include <string.h>
 #include <iostream>
 #include <algorithm>
 #include <vector>
 #include <queue>
 #include <set>
 #include <map>
 #include <string>
 #include <math.h>
 #include <stdlib.h>
 #include <time.h>
 using namespace std;
 typedef unsigned long long ull;
 const int maxn = 1e4 + ;
 const int seed = ;
 ull HASH, s[maxn*], p[maxn*];
 char a[maxn], b[maxn*], ans[maxn*];
 void init() {
     p[] = ;
     for (int i =  ; i < maxn ; i++)
         p[i] = p[i - ] * seed;
 }
 int main() {
     init();
     int t;
     scanf("%d", &t);
     while(t--) {
         scanf("%s%s", a, b );
         int lena = strlen(a), lenb = strlen(b);
         HASH = ;
         for (int i =  ; i < lena ; i++)
             HASH = HASH * seed + a[i];
         int top = ;
         s[] = ;
         int sum = ;
         for (int i =  ; i < lenb ; i++) {
             ans[top++] = b[i];
             s[top] = s[top - ] * seed + b[i];
             if ( top >= lena && s[top] - s[top - lena]*p[lena] == HASH ) sum++;
         }
         printf("%d\n", sum);
     }
     return ;
 }

外加一个kmp写法吧  毕竟多学点东西肯定没有错

 #include <stdio.h>
 #include <string.h>
 #include <iostream>
 #include <algorithm>
 #include <vector>
 #include <queue>
 #include <set>
 #include <map>
 #include <string>
 #include <math.h>
 #include <stdlib.h>
 #include <time.h>
 using namespace std;
 typedef unsigned long long ull;
 const int maxn = 1e4 + ;
 int nxt[maxn], lena, lenb, ans;
 char a[maxn], b[maxn * ];
 void Get_nxt() {
     int i = , j = - ;
     nxt[] = -;
     while(i < lena) {
         if (j == - || a[i] == a[j] ) {
             ++i, ++j;
             nxt[i] = j;
         } else j = nxt[j];
     }
 }
 void kmp() {
     int i = , j =  ;
     Get_nxt();
     while(i < lenb) {
         if (j == - || b[i] == a[j]) {
             ++i, ++j;
             if (j == lena) ans++;
         } else j = nxt[j];
     }
 }
 int main() {
     int t;
     scanf("%d", &t);
     while(t--) {
         ans = ;
         scanf("%s%s", a, b);
         lena = strlen(a);
         lenb = strlen(b);
         kmp();
         printf("%d\n", ans);
     }
     return ;
 }