【POJ】3461Oulipo（字符串hash例题）

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 53276		Accepted: 20956

Description

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

Source

BAPC 2006 Qualification

题目大意：给你两个字符串，S1,S2，问你S1在S2中出现的次数，两次出现可以有重复的点，

思路：这个题目有两种做法吧，一种是KMP，或者扩展KMP做法吧，还有一种就是现在要说的字符串hash

所谓字符串hash，在我看来，基本上就是把一个字符串看成一个一元多次的函数，几次的就看他的长度了，字符串的每个字母的ASCII值作为每一项的数值系数，类似于下面这样子的

然后取一个基本无重复的b最后对比数值即可

代码：

#include
#include
#include
#include
#define ULL unsigned long long
using namespace std;
const int maxn=1000000+10;
const ULL b=100000000+7;
ULL power[maxn];
ULL sum[maxn];
void init()
{
    power[0]=1;
    for(int i=1; i<=maxn; i++)
        power[i]=power[i-1]*b;
}
int main()
{
    int t;
    init();
    scanf("%d",&t);
    while(t--)
    {
        char s1[maxn],s2[maxn];
        scanf("%s%s",s1+1,s2+1);
        int n=strlen(s1+1),m=strlen(s2+1);
        memset(sum,0,sizeof(sum));
        sum[0]=0;
        for(int i=1; i<=m; i++)
        {
            sum[i]=sum[i-1]*b+(ULL)(s2[i]-'A'+1);
        }
        ULL s=0;
        for(int i=1; i<=n; i++)
        {
            s=s*b+(ULL)(s1[i]-'A'+1);
        }
        int ans=0;
        for(int i=0; i<=m-n; i++)
            if(s==sum[i+n]-sum[i]*power[n])
            {
                ans++;
            }
        printf("%d\n",ans);
    }
}