poj1200 Crazy Search

Description Many people like to solve hard puzzles some of which may
lead them to madness. One such puzzle could be finding a hidden prime
number in a given text. Such number could be the number of different
substrings of a given size that exist in the text. As you soon will
discover, you really need the help of a computer and a good algorithm
to solve such a puzzle. Your task is to write a program that given the
size, N, of the substring, the number of different characters that may
occur in the text, NC, and the text itself, determines the number of
different substrings of size N that appear in the text.

As an example, consider N=3, NC=4 and the text “daababac”. The
different substrings of size 3 that can be found in this text are:
“daa”; “aab”; “aba”; “bab”; “bac”. Therefore, the answer should be 5.

Input The first line of input consists of two numbers, N and NC,
separated by exactly one space. This is followed by the text where the
search takes place. You may assume that the maximum number of
substrings formed by the possible set of characters does not exceed 16
Millions.

Output The program should output just an integer corresponding to the
number of different substrings of size N found in the given text.

字符串哈希。
写起来很难受。
乱搞一下就好了。

#include<cstdio>
#include<cstring>
char s[1000010];
bool hash[16000010];
int pow[10010],s1[1000010],map[1010];
int main()
{
    int i,j,k,l,m,n,p,q,x,y,z,ans,cnt;
    while (scanf("%d%d",&k,&p)==2)
    {
        cnt=0;
        pow[0]=1;
        for (i=1;i<=k;i++)
          pow[i]=pow[i-1]*p;
        memset(hash,0,sizeof(hash));
        memset(map,0xff,sizeof(map));
        scanf("%s",s+1);
        l=strlen(s+1);
        if (l<k)
        {
            printf("0\n");
            continue;
        }
        for (i=1;i<=l;i++)
        {
            if (map[s[i]]==-1) map[s[i]]=cnt++;
            s1[i]=map[s[i]];
        }
        x=0;
        for (i=1;i<=k;i++)
          x=x*p+s1[i];
        hash[x]=1;
        ans=1;
        for (i=k+1;i<=l;i++)
        {
            x-=s1[i-k]*pow[k-1];
            x*=p;
            x+=s1[i];
            if (!hash[x]) ans++;
            hash[x]=1;
        }
        printf("%d\n",ans);
    }
}