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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.
Sample Input 3 4 daababac Sample Output 5 Hint Huge input,scanf is recommended.
Source |
这一题一看就是很大的数据量，从前往后接着比较每个字符串是否出现过。
刚开始我用ELF想HASH字符串，无奈ELF碰撞的太多，而且还要存储下字符串便于比较，餐具啊，这个题的数据量太大，赵超不误
就思考了，也看了别人说的用NC进制HASH，易证，可以保持一一对应，而不出现碰撞，这所以只需要一个数组就够了。
完美啊
接着我就从前往后找，更新的时候只需要把前面那个减去加上新加上的那个，再用hash比较，即可
16MS，C++提交，犀利
#include<stdio.h>
#include<string.h>
const int N=16000010;
char str[N];
bool hash[N];
int a[10000];
int main()
{
int n,nc;
while(scanf("%d%d",&n,&nc)==2)
{
memset(a,0,sizeof(a));
memset(hash,0,sizeof(hash));
scanf("%s",str);
int l=strlen(str);
int num=0;
for(int i=0;i<l;i++)
{
if(!a[str[i]]) a[str[i]]=num++;
}
int sum=0,t=0,bb=1;
for(int i=0;i<n;i++)
t=t*nc+a[str[i]];
if(hash[t]==0)
{
sum++;
hash[t]=true;
}
for(int i=1;i<n;i++) bb=bb*nc;
for(int i=0;i<=l-n-1;i++)
{
t=(t-a[str[i]]*bb)*nc+a[str[i+n]];
if(hash[t]==0)
{
sum++;
hash[t]=true;
}
}
printf("%d/n",sum);
}
return 0;
}
