MUH and Cube Walls

D - MUH and Cube Walls
Crawling in process...Crawling failedTime
Limit:2000MS    Memory Limit:262144KB    
64bit IO Format:%I64d & %I64u
SubmitStatus

Description

Input

Output

Sample Input

Sample Output

Hint

Description

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 ofw towers. The bears also finished making their wall but they didn't give it a name. Their wall consists ofn
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 ofw 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".

Input

The first line contains two integers n andw (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.

Output

Print the number of segments in the bears' wall where Horace can "see an elephant".

Sample Input

Input

13 5
2 4 5 5 4 3 2 2 2 3 3 2 1
3 4 4 3 2

Output

2

Sample Output

Hint

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.



题意：给你一个长度为n（1<=n<=2e5）的一排积木，长度分别为a[]（1<=a[]<=1e9）。 
有一个长度为m（1<=m<=2e5）的一排积木，长度分别为b[]（1<=b[]<=1e9）  
问你，第一排积木有多少个位点i，使得[i+0,i+m-1]这一段积木，之间增减幅度与b[]的整体增减幅度相同。 增减幅度肯定产生于相邻的积木之间。 
于是我们求出第二排积木之间的m-1个增减幅度。 然后求出第一排积木之间的n-1个增减幅度。 然后以第二个串作为模板串，第一个串为匹配串。 做KMP匹配求匹配位点个数

#include<stdio.h>
#include<iostream>
#include<string.h>
#include<string>
#include<ctype.h>
#include<math.h>
#include<algorithm>
#include<time.h>
using namespace std;
typedef long long LL;
typedef unsigned long long UL;
typedef unsigned int UI;
const int N=2e5+10,M=0,Z=1e9+7,ms63=1061109567;
int n,m;
int w
,a
,b
;
int we
;
void next()
{
int j=0;we[1]=0;
for(int i=2;i<=m;++i)
{
while(j&&b[j+1]!=b[i])j=we[j];
if(b[j+1]==b[i])++j;
we[i]=j;
}
}
void kmp()
{
int ans=0;
int j=0;
for(int i=1;i<=n;++i)
{
while(j&&b[j+1]!=a[i])j=we[j];
if(b[j+1]==a[i])++j;
if(j==m)++ans;
}
printf("%d\n",ans);
}
int main()
{
while(~scanf("%d%d",&n,&m))
{
for(int i=1;i<=n;++i)scanf("%d",w[i]);
for(int i=1;i<n;++i)a[i]=w[i]-w[i+1];
for(int i=1;i<=m;++i)scanf("%d",&w[i]);
for(int i=1;i<m;++i)b[i]=w[i]-w[i+1];
if(m==0)
{
printf("%d\n",n+1);
continue;
}
next();
kmp();
}
return 0;
}