codeforces 471D MUH and Cube Walls kmp

题目链接:cf 471d

D. MUH and Cube Walls

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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".

Input

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.

Output

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

Sample test(s)

input

13 52 4 5 5 4 3 2 2 2 3 3 2 13 4 4 3 2

output

2

Note

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.

题目大意：给定两个数组表示形状,问第二个数组的形状能否在第一个数组中找到.若能找到,求出能找到几次                   kmp,因为只需要形状相同,不需要考虑每个值的实际大小,只需要使得数组的前后差值相等,就能使得上边界构成的形状相同。        求出两个数组的相邻数差值构成的新数组,然后利用kmp求第二个差值数组在第一个差值数组中出现的次数,注意到若第二个数组就只有一个数,则在第一个数组任意位置都能构成相同的形状因此此时输出n即可

/****************************************************** * File Name:   d.cpp * Author:      kojimai * Creater Time:2014年09月27日 星期六 00时19分37秒 ******************************************************/#include<cstdio>#include<cstring>#include<cmath>#include<algorithm>#include<iostream>using namespace std;#define FFF 200005long long a[FFF],b[FFF];long long c[FFF],d[FFF];int next[FFF];int main(){int n,w;cin>>n>>w;for(int i=0;i<n;i++){cin>>a[i];if(i)c[i-1]=a[i]-a[i-1];}for(int i=0;i<w;i++){cin>>b[i];if(i)d[i-1]=b[i]-b[i-1];}if(w==1){cout<<n<<endl;}//dgbug/*for(int i=0;i<n-1;i++)  cout<<c[i]<<' ';cout<<endl;  for(int i=0;i<w-1;i++)  cout<<d[i]<<' ';cout<<endl; */else{next[0]=-1;int index=0;next[1]=0;for(int i=2;i<w;i++){while(index>=0&&d[index]!=d[i-1])index=next[index];next[i]=++index;}int ans=0;int l1=0,l2=0;while(l1<n-1){if(c[l1]==d[l2]||l2==-1){l1++;l2++;if(l2==w-1){ans++;l2=next[l2];}}elsel2 = next[l2];}cout<<ans<<endl;}return 0;}