Oulipo HDU

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 + 10;
const int seed = 13331;
ull HASH, s[maxn*100], p[maxn*100];
char a[maxn], b[maxn*100], ans[maxn*100];
void init() {
    p[0] = 1;
    for (int i = 1 ; i < maxn ; i++)
        p[i] = p[i - 1] * seed;
}
int main() {
    init();
    int t;
    scanf("%d", &t);
    while(t--) {
        scanf("%s%s", a, b );
        int lena = strlen(a), lenb = strlen(b);
        HASH = 0;
        for (int i = 0 ; i < lena ; i++)
            HASH = HASH * seed + a[i];
        int top = 0;
        s[0] = 0;
        int sum = 0;
        for (int i = 0 ; i < lenb ; i++) {
            ans[top++] = b[i];
            s[top] = s[top - 1] * seed + b[i];
            if ( top >= lena && s[top] - s[top - lena]*p[lena] == HASH ) sum++;
        }
        printf("%d
", sum);
    }
    return 0;
}

 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 + 10;
int nxt[maxn], lena, lenb, ans;
char a[maxn], b[maxn * 100];
void Get_nxt() {
    int i = 0, j = -1 ;
    nxt[0] = -1;
    while(i < lena) {
        if (j == -1 || a[i] == a[j] ) {
            ++i, ++j;
            nxt[i] = j;
        } else j = nxt[j];
    }
}
void kmp() {
    int i = 0, j = 0 ;
    Get_nxt();
    while(i < lenb) {
        if (j == -1 || b[i] == a[j]) {
            ++i, ++j;
            if (j == lena) ans++;
        } else j = nxt[j];
    }
}
int main() {
    int t;
    scanf("%d", &t);
    while(t--) {
        ans = 0;
        scanf("%s%s", a, b);
        lena = strlen(a);
        lenb = strlen(b);
        kmp();
        printf("%d
", ans);
    }
    return 0;
}