題目連結：

http://172.16.0.132/senior/#main/show/5437

題目：

題解：

發現滿足上述性質並且僅當A序列的子序列的差分序列與B序列的差分序列相同

於是我們把A變成差分序列，把B變成差分序列，做一次KMP就好了

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

const int N=1e6+15;
int n,m;
int a[N],b[N],nxt[N];
inline 
 
int read(){
    char ch=getchar();int s=0,f=1;
    while (ch<'0'||ch>'9') {if (ch=='-') f=-1;ch=getchar();}
    while (ch>='0'&&ch<='9') {s=(s<<3)+(s<<1)+ch-'0';ch=getchar();}
    return s*f;
}
int main(){
    freopen("sequence.in","r",stdin);
    freopen("sequence.out
 
","w",stdout);
    n=read();m=read();
    for (int i=1;i<=n;i++) a[i]=read();
    for (int i=1;i<=m;i++) b[i]=read();
    for (int i=1;i<n;i++) a[i]=a[i+1]-a[i];
    for (int i=1;i<m;i++) b[i]=b[i+1]-b[i];
    n--;m--;
    nxt[1]=0;
    for (int i=2,j=0;i<=m;i++){
        while
 
 (j&&b[j+1]!=b[i]) j=nxt[j];
        if (b[j+1]==b[i]) ++j;
        nxt[i]=j;
    }
    int ans=0;
    for (int i=1,j=0;i<=n;i++){
        while (j&&(j==m||b[j+1]!=a[i])) j=nxt[j];
        if (b[j+1]==a[i]) ++j;
        if (j==m) ans++;
    }
    printf("%d\n",ans);
    return 0;
}