【字符串Hash模板题】 POJ - 3461 A - Oulipo

A - Oulipo  POJ - 3461 

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.

Input

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.

Output

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.

Sample Input

3
BAPC
BAPC
AZA
AZAZAZA
VERDI
AVERDXIVYERDIAN

Sample Output

1
3
0

给你一个子串和一个父串，问你父串中有几个子串

1、直接ah[i]=ah[i-1]*base+a[i]，那么长度为len的哈希值就是ah[i]-ah[i-len]*base[len]  此处ah[i]表示长为i字符串的哈希值

北京邀请赛的题，为了能做到在O(1)时间内算出hash值---比如算从l到r这段字符串的hash值，可以直接ah[r]-ah[l-1]（当然需要判断l>=1，或者直接字符串从下标为1开始），便于进行二分，就是这么写的

（http://blog.csdn.net/u011026968/article/details/38473707）

2、逐步递推：hash[i]=hash[i-1]*base+a[i]-a[i-len]*base[len]  三、HASH的其他构造方法：
1、如果限定了字符数目，可以映射为相应进制的数（http://blog.csdn.net/u011026968/article/details/38490653）

#include 
#include 
#include 
#define ll long long
using namespace std;
const int maxn=1000005;
const int base=131;
char son[maxn],fa[maxn];
ll p[maxn],hash[maxn];

int main()
{
    int T;
    scanf("%d",&T);
    p[0]=1;
    for(int i=1;i=0;i--)
        {
            son_num=son_num*base+son[i];   //得到子串的hash值
        }

        hash[R]=0;
        for(int i=R-1;i>=0;i--)
        {
            hash[i]=hash[i+1]*base+fa[i];  //保存每个位置时的父串的hash值
        }

        int ans=0;
        for(int i=0;i<=R-L;i++)
        {
            if(son_num==hash[i]-hash[i+L]*p[L])  //由公式得到hash[r-l]的值.
                ans++;
        }
        printf("%d\n",ans);
    }
    return 0;
}