[Ybtoj Chapter 8 Example 4] Substring splitting [KMP]

---

Problem-solving ideas

——Because an array has been reduced and adjusted for 1 hour, I am speechless...TOT

Because the data is not big for this question, you can run n 2 n^2 directly$n^2$ .
Consider enumerating the left end point, let the left end point be$l，S=C[l,|C|]$ , then runKMP KMPon the string S$KMP$ .
Then in the statistical process, assuming that it is now at position i, then$S[1,j]=S[i-j+1,i]$。如果$j\ast 2>=i$ , then let j=p[j], at this time$A[1,j]=a[i-j+1,i],j$ keeps jumping back along the pointer until$j\ast 2<i$ . Then judge whether j is greater than or equal to k, and if so, add up the answers.

---

Code

#include<iostream>
#include<cstdio>
#include<iomanip>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;

char s[20010],c[20010];
int ans,j,k,l,len,m,p[20010];

int main() {
    
    
	scanf("%s",c+1);
	scanf("%d",&k);
	len=strlen(c+1);
	for(register int t=1; t<=len; t++) {
    
    
		memset(s,0,sizeof(s));
		memset(p,0,sizeof(p));
		for(register int w=1;w+t-1<=len;w++)
			s[w]=c[w+t-1]; //处理左端点
		j=0,l=strlen(s+1);
		for(register int i=2; i<=l; i++) {
    
    
			while(j&&s[i]!=s[j+1])
				j=p[j];
			if(s[i]==s[j+1])
				j++;
			p[i]=j;
		}
		j=0;
		for(register int i=1; i<=l; i++) {
    
    
			while(j&&s[i]!=s[j+1])
				j=p[j];
			if(s[i]==s[j+1])
				j++;
			while((j<<1)>=i)j=p[j];
			if(j>=k)ans++;
		}
	}
	printf("%d",ans);
}