Extended Los KMP Valley P5410 template

Topic links: https://www.luogu.org/problem/P5410

Question is intended: to two strings a, b, b seeking a maximum prefix length for each of the suffixes a

Analysis: Extended KMP (also known as Z-algorithm algorithm) bare title

The blog to explain better: https://www.luogu.org/blog/lc-2018-Canton/solution-p5410

But there are several places he talked about a few problems, mainly in the case 2 inside

First, a start S [K + L] is definitely the back of p, the obvious

Case 2 red blue and green and then sequence the three lines is not the same, red and green is still the same, but the blue line will not wait, take a look at this myself

#include <bits / STDC ++ H.>
 the using  namespace STD; 
typedef Long  Long LL;
 const  int MAXN = + 1E5 . 7 ; // interword add their own symbols, slightly enlarged 
const LL INF = 1e18;
 #define meminf (A) Memset (A, 0x3F, the sizeof (A))
 #define MEM0 (A) Memset (A, 0, the sizeof (A));
 char A [MAXN], B [MAXN];
 int NXT [MAXN], Extend [MAXN]; // NXT [I] Representative b [i ... len] and b is the maximum prefix length, extend [i] representative of a [i ... len] and a maximum prefix length b 
void getnxt () {
     int len = strlen (B); 
    NXT [ 0 ] = len; //[0] is the nxt original strings case, naturally, for the len 
    int now = 0 ;
     the while (B [now] == B [now + . 1 ] && now + . 1 <len) now ++ ; 
    nxt [ . 1 ] = now;
     int P0 = . 1 ; // P0 is satisfied p0 nxt [p0] -1 + maximum point, the beginning of initialization. 1 
    for ( int I = 2 ; I <len; I ++ ) {
         IF (I + NXT [P0-I] <+ P0 NXT [P0]) NXT [I] = NXT [I- P0];
         // I corresponds to k + 1, nxt [i- p0] corresponds to L (nxt [k-p0 + 2) where the string is from 0 started, and the difference from a blog which is beginning a small increase 1 
        the else {
             int now NXT = P0 + [P0] -  I;
            now=max(now,0);//防止i在p的后面
            while(b[now]==b[i+now]&&i+now<len) now++;
            nxt[i]=now;
            p0=i;//更新p0 
        } 
    } 
}
void exkmp(){
    getnxt();
    int len=strlen(a);
    int now=0;
    while(a[now]==b[now]&&now<min(strlen(a),strlen(b))) now++;
    extend[0]=now;
    int p0=0;
    for(int i=1;i<len;i++){
        if(i+nxt[i-p0]<extend[p0]+p0) extend[i]=nxt[i-p0];
        else{
            int now=p0+extend[p0]-i;
            now=max(now,0);
            while(b[now]==a[i+now]&&now<strlen(b)&&i+now<len) now++;
            extend[i]=now;
            p0=i;
        }
    }  
}
int main(){
    scanf("%s%s",a,b);
    exkmp();
    for(int i=0;i<strlen(b);i++) printf("%d ",nxt[i]);
    putchar('\n');
    for(int i=0;i<strlen(a);i++) printf("%d ",extend[i]);
    putchar('\n');
    return 0;
}