HUD1686Oulipokmp模板题/哈希模板题

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


题意：计算模式串在原串中出现的次数

对于next数组的理解是理解kmp的关键。
next[i]:记录的是前后缀最长公共长度。

KMP：

#include<stdio.h>
#include<iostream>
#include<string.h>
using namespace std;

const int N=1000020;
const int M=10020;

char s[N];//原串
char t[M];//模式串
int nextt[M];

void getnext(int len)//求的是模式串的next数组
{
    int i=0,j=-1;
    nextt[0]=-1;
    while(i<len)
    {
        if(j<0||t[i]==t[j])
            nextt[++i]=++j;
        else
            j=nextt[j];
    }
}

int kmp(int m,int n)//m模式串长度、n原串长度
{
    int i=0,j=0,ans=0;
    while(i<n)
    {
        if(j==-1||t[j]==s[i])
        {
            i++;
            j++;
        }
        else
            j=nextt[j];
        if(j==m)
        {
            ans++;
            j=nextt[j];
        }
    }
    return ans;
}

int main()
{
    int tt;
    scanf("%d",&tt);
    while(tt--)
    {
        memset(s,'\0',sizeof(s));
        memset(t,'\0',sizeof(t));
        memset(nextt,0,sizeof(nextt));
        scanf("%s%s",t,s);//模式串、原串
        int len1=strlen(t);//模式串
        int len2=strlen(s);//原串
        getnext(len1);
        printf("%d\n",kmp(len1,len2));
    }
    return 0;
}

哈希：

#include<stdio.h>
#include<iostream>
#include<cmath>
#include<algorithm>
#include<string.h>
#include<queue>
#include<map>
using namespace std;
typedef unsigned long long ull;
const int N=1e6+20;

char a[N],b[N];
ull p[N],sum[N],x=131;
//求a(子串)在b(母串)中出现多少次

void w()
{
    p[0]=1;
    for(int i=1; i<1000000; i++)
        p[i]=p[i-1]*x;//预处理出x^n
}

int main()
{
    w();
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%s %s",a+1,b+1);//使得下标从1开始
        int la=strlen(a+1);//短
        int lb=strlen(b+1);//长
        sum[0]=0;
        for(int i=1; i<=lb; i++)
            sum[i]=sum[i-1]*x+(ull)(b[i]-'A'+1);
        ull s=0;
        for(int i=1; i<=la; i++)
            s=s*x+(ull)(a[i]-'A'+1);//*x是为了化成x进制数
        int ans=0;
        for(int i=0; i<=lb-la; i++)
        {
            if(s==sum[i+la]-sum[i]*p[la])
                ans++;
        }
        printf("%d\n",ans);
    }
    return 0;
}

不明白为什么哈希的时间比kmp慢而且占用的内存都快是kmp的10倍了？？？