Codeforces 471D MUH and Cube Walls【思维+KMP】

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 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 containsn integers a_i (1 ≤ a_i ≤ 10⁹) — the heights of the towers in the bears' wall. The third line contains w integers b_i (1 ≤ b_i ≤ 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".

Examples

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.

问题比较抽象，我们对于题目大意进行简化说明：

给你一个数组A，再给你一个数组B，我们可以任意将数组B整体增加或者减少值X（整数），我们可以进行修改值的操作无限次，每次我们要在数组A中找寻有几段和数组B完全匹配的子段。

思路：

如果我们枚举这个修改值X的话，时间复杂度非常的高，这里我们只要想到了差值，那么这个题就结束了。

对应数组B，我们无论对于整体修改值X为多大，其差值都是一样的：

3 4 4 3 2

-1 0 1 1 

所以我们只要对于这个差值进行KMP匹配即可。

当M==1的时候需要特判，答案为N.

Ac代码：

#include<stdio.h>#include<string.h>using namespace std;int tmpa[2500050];int tmpb[2500050];int a[250050];int b[250050];int next[250050];int n,m;int lena;int lenb;int output;void set_naxt()//子串的next数组{    int i=0,j=-1;    next[0]=-1;    while(i<lenb)    {        if(j==-1||b[i]==b[j])        {            i++; j++;            next[i]=j;        }        else        j=next[j];    }}int kmp(){    int i=0,j=0;    set_naxt();    while(i<lena)    {        if(j==-1||a[i]==b[j])        {            i++;j++;        }        else        j=next[j];        if(j==lenb)        {            output++;            j=next[j];            //printf("%d\n",j);        }    }    return -1;}int main(){    while(~scanf("%d%d",&n,&m))    {        for(int i=0;i<n;i++)scanf("%d",&tmpa[i]);        for(int i=0;i<m;i++)scanf("%d",&tmpb[i]);        if(m==1)        {            printf("%d\n",n);            continue;        }        for(int i=0;i<n-1;i++)tmpa[i]=tmpa[i]-tmpa[i+1];        for(int i=0;i<m-1;i++)tmpb[i]=tmpb[i]-tmpb[i+1];        lena=n-1;        lenb=m-1;        for(int i=0;i<n-1;i++)a[i]=tmpa[i];        for(int i=0;i<m-1;i++)b[i]=tmpb[i];        output=0;        kmp();        printf("%d\n",output);    }}