POJ-1200-Crazy Search（字符串Hash）

Crazy Search

Time Limit: 1000MS Memory Limit: 65536K

Total Submissions: 33142 Accepted: 9079

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 4

daababac

Sample Output

分析

串的长度最大为16000000，所以如果要用字符串Hash（把字符串当作一个数字）就要尽量使这个数字小。
利用字符在串中的出现顺序当作该字符的权值，而整个字符串就是nc进制数，然后把数字用hash表储存。

const int p = 131;

char has[16000001];

int a[128];

string str;

int n,nc;

int num = 0,cnt = 0;

int main()

{

	cin>>n>>nc;

	cin>>str;

	int len = str.length();

	for(int i=0;i<len;i++)

	{

		if(!a[str[i]])

			a[str[i]] = ++num;

	}

	for(int i=0;i<len-n+1;i++)

	{

		int sum=0;

		for(int j=i;j<i+n;j++)

			sum = sum*nc+a[str[i]];

		if(!has[sum])

			has[sum] = 1,cnt++;

	}

	cout<<cnt<<endl;

}

巴特西

POJ-1200-Crazy Search（字符串Hash）

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