Crazy Search POJ

Crazy Search

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 29481 Accepted: 8191

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.

Sample Input

3 4daababac

Sample Output

5

Hint

Huge input,scanf is recommended.

题目分析：给一个一个长的串，然后求其中有多少个长度为n的不同字串。因为题目中给出了nc，即字母种类的个数。所以可以把每一个出现的字母对应为1到nc。然后把长度为n的字符串映射成为一个整数。即把这个字符串看成一个nc进制的数。

代码：

/* Author:kzl */#include<iostream>#include<cstring>#include<algorithm>#include<cstdio>using namespace std;typedef long long LL;const int maxx = 16000000+7;int ha[maxx];char ch[maxx];int ma[300];int n,m;int main(){while(scanf("%d%d",&n,&m)!=EOF){    scanf("%s",ch);    int len = strlen(ch);    LL ans = 0;    memset(ha,0,sizeof(ha));    memset(ma,0,sizeof(ma));    int num = 0;    for(int i=0;i<len;i++)    {        if(ma[ch[i]]==0)ma[ch[i]] = ++num;        if(num==m)break;    }    for(int i=0;i<len - n+1;i++)    {        LL sum = 0;        for(int j=i;j<i+n;j++){        sum = sum * m + ma[ch[j]];        }        if(!ha[sum])ha[sum] = 1,ans++;    }    printf("%lld\n",ans);}return 0;}