HUD1686-Oulipo-kmp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=;
 const int M=;

 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=,j=-;
     nextt[]=-;
     while(i<len)
     {
         if(j<||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=,j=,ans=;
     while(i<n)
     {
         if(j==-||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,,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 ;
 }

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+;

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

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

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

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？？？