Polar bears Menshykov and Uslada from the zoo of St. Petersburg and elephant Horace from the zoo of Kiev got hold of lots of wooden cubes somewhere. They started making cube towers by placing the cubes one on top of the other. They defined multiple towers standing in a line as a wall. A wall can consist of towers of different heights.
Horace was the first to finish making his wall. He called his wall an elephant. The wall consists of w towers. The bears also finished making their wall but they didn't give it a name. Their wall consists of n towers. Horace looked at the bears' tower and wondered: in how many parts of the wall can he "see an elephant"? He can "see an elephant" on a segment of w contiguous towers if the heights of the towers on the segment match as a sequence the heights of the towers in Horace's wall. In order to see as many elephants as possible, Horace can raise and lower his wall. He even can lower the wall below the ground level (see the pictures to the samples for clarification).
Your task is to count the number of segments where Horace can "see an elephant".
The first line contains two integers n and w (1 ≤ n, w ≤ 2·105) — the number of towers in the bears' and the elephant's walls correspondingly. The second line contains n integers ai (1 ≤ ai ≤ 109) — the heights of the towers in the bears' wall. The third line contains w integers bi (1 ≤ bi ≤ 109) — the heights of the towers in the elephant's wall.
Print the number of segments in the bears' wall where Horace can "see an elephant".
13 5 2 4 5 5 4 3 2 2 2 3 3 2 1 3 4 4 3 2
2
The picture to the left shows Horace's wall from the sample, the picture to the right shows the bears' wall. The segments where Horace can "see an elephant" are in gray.
先计算出两个数组的相邻两个数之差，
利用kpm匹配，计算出子串数量。
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <iomanip>
#include <algorithm>
#include <climits>
#include <cstring>
#include <string>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <vector>
#include <list>
#define rep(i,m,n) for(int i=m;i<=n;i++)
#define rsp(it,s) for(set<int>::iterator it=s.begin();it!=s.end();it++)
const int inf_int = 2e9;
const long long inf_ll = 2e18;
#define inf_add 0x3f3f3f3f
#define mod 1000000007
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define pi acos(-1.0)
#define pii pair<int,int>
#define Lson L, mid, rt<<1
#define Rson mid+1, R, rt<<1|1
const int maxn=5e2+10;
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
inline int read(){int ra,fh;char rx;rx=getchar(),ra=0,fh=1;
while((rx<'0'||rx>'9')&&rx!='-')rx=getchar();if(rx=='-')
fh=-1,rx=getchar();while(rx>='0'&&rx<='9')ra*=10,ra+=rx-48,
rx=getchar();return ra*fh;}
//#pragma comment(linker, "/STACK:102400000,102400000")
ll gcd(ll p,ll q){return q==0?p:gcd(q,p%q);}
ll qpow(ll p,ll q){ll f=1;while(q){if(q&1)f=f*p;p=p*p;q>>=1;}return f;}
typedef vector<int> vi;
int dir[4][2]={{-1,0},{1,0},{0,-1},{0,1}};
const int N = 1e5+5;
int n,w;
vi a,b;
ll aa[N*2];
ll bb[N*2];
void kmp_pre(ll x[],int m,int next[])
{ int i,j; j=next[0]=-1; i=0;
while(i<m)
{
while(-1!=j && x[i]!=x[j])
j=next[j]; next[++i]=++j;
}
}
int KMP_Count(ll x[],int m,ll y[],int n)
{
int next1[N*2];
memset(next1,0,sizeof(next1));
int i,j;
int ans=0;
kmp_pre(x,m,next1);
i=j=0;
while(i<n)
{
while(-1!=j && y[i]!=x[j]) j=next1[j];
i++;j++;
if(j>=m)
{ ans++; j=next1[j]; }
}
return ans;
}
int main()
{
int t;
cin >> n>>w;
rep(i,1,n)
{
cin>>t;
a.push_back(t);
}
rep(i,1,w)
{
cin >> t;
b.push_back(t);
}
if(w==1)
{
cout << n <<endl;
return 0;
}
for(int i=0;i<a.size()-1;i++){
aa[i] =a[i+1]-a[i];
}
for(int i=0;i<b.size()-1;i++){
bb[i] =b[i+1]-b[i];
}
cout <<KMP_Count(bb,w-1,aa,n-1)<<endl;
return 0;
}