Crazy Search (哈稀)_zoj crazy search-优快云博客

本文介绍了一种使用哈希思想解决寻找文本中特定大小的不同子串数量的问题，通过将字符映射为整数并计算哈希值来高效地找出所有可能的不同子串。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

题目链接：https://vjudge.net/contest/247842#problem/B

参考链接：https://blog.youkuaiyun.com/qq_38538733/article/details/77340299

Crazy Search

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

Hint

Huge input,scanf is recommended.

哈稀的思想，就是一个映射的过程，

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
const int MAXN=16000006;
int  has[MAXN];
char s[MAXN];
int  vis[500];

int main(){
    int n,m;
    while(cin>>n>>m){
        scanf("%s",s+1);

        //涨知识了，memset函数还能导致内存超限
        memset(vis,0,sizeof(vis));
        memset(s,0,sizeof(s));
        memset(has,0,sizeof(has));


        int len=strlen(s+1);
        memset(vis,0,sizeof(vis));

        int num=0;
        for(int i=1;i<=len;i++){
            if(!vis[s[i]]){
                vis[s[i]]=num++;//把每一个不重复的字母都赋一个数字值，从0 ～ 最多26
            }
        }
        int ans=0;
        for(int i=1;i<=len-n+1;i++){
            int sum=0;
            for(int j=0;j<n;j++){
                sum=sum*num+vis[s[i+j]];//将从i~j的n个长度的字符串转换成数字
            }
            //此时sum的值代表n长度的字符串转换成数字的值
            if(!has[sum]){
                has[sum]=1;
                ans++;
            }
        }
        printf("%d\n",ans);

    }

    return 0;
}