Poj 1200 (hash)

Crazy Search

Time limit:1000 ms		Memory limit:65536 K
Total submissions:23168		Accepted:6513

Description

Please people like to solve hard puzzles some of which may lead them to madness. one such puzzle cocould be finding a hidden prime number in a given text. such number cocould 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 "daabac ". the different substrings of size 3 that can be found in this text are: "Daa"; "AAB"; "ABA"; "Bab"; "BAC ". therefore, the answer shoshould 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 shocould output just an integer corresponding to the number of different substrings of size N found in the given text.

Sample Input

3 4daababac

Sample output

5

Hint

Huge input, scanf is recommended.

Source

Southwestern Europe 2002

Question: Enter n, m, and the number of characters in the string (M). How many types of strings are there?

Hash. Check the code.

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<map>using namespace std;#define N 16000005char a[N];int n,m;int ha[N];int main(){    int i,j;    while(~scanf("%d%d",&n,&m))    {        scanf("%s",a);        i=0;        memset(ha,0,sizeof(ha));        int len=strlen(a);        for(j=0;j<len;j++)        {            if(!ha[a[j]])            {                ha[a[j]]=++i;                if(i==m)                    break;            }        }        int ans=0;        for(i=0;i+n<=len;i++)        {            int temp=0;            for(j=i;j<i+n;j++)            {                temp=temp*m+ha[a[j]];            }            if(!ha[temp])            {                ans++;                ha[temp]=1;            }        }        printf("%d\n",ans);        return 0;    }    return 0;}

Poj 1200 (hash)