CodeForces 471D MUH and Cube Walls (KMP + 差分)

D. MUH and Cube Walls

time limit per test: 2 seconds
memory limit per test: 256 megabytes
input: standard input
output: standard output

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 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 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 5
2 4 5 5 4 3 2 2 2 3 3 2 1
3 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.

Solution

思路

首先差分，计算出主串和模式串中相邻数字的差分数组。(去掉首位的)

之后就是 KMP 匹配这两个差分数组了了。

AC 代码

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <set>
#include <map>
#include <queue>
#include <vector>
#include <utility>
#include <algorithm>
#define MAXN 200005
#define INF 0x3f3f3f3f
#define DEBUG
#define DataIn
typedef long long LL;

using namespace std;

inline int read()
{
    int x=0,f=1;char ch=getchar();
    while(!isdigit(ch)) {if(ch=='-') f=-1;ch=getchar();}
    while(isdigit(ch)) {x=x*10+ch-'0';ch=getchar();}
    return x*f;
}

int n, w;
int a[MAXN], b[MAXN];
int next[MAXN];

void prekmp()
{
    int i=0, j=-1;
    next[0] = -1;
    while(i<w)
    {
        while(~j && b[i]!=b[j])
            j = next[j];
        next[++i] = ++j;
    }
}

int kmpcnt()
{
    int i=0, j=0;
    int cnt = 0;
    prekmp();
    while(i<n)
    {
        while(~j && a[i]!=b[j])
            j = next[j];
        i++; j++;
        if(j >= w)
        {
            cnt++;
            j = next[j];
        }
    }
    return cnt;
}

int main()
{
    n = read(), w = read();
    for(int i=0; i<n; i++)
        a[i] = read();
    for(int i=n-1; i>0; i--)
        a[i] -= a[i-1];
    for(int i=0; i<n-1; i++)
        a[i] = a[i+1];
//    for(int i=0; i<n-1; i++)
//        printf("%d ", a[i])
//    putchar(10);
    n--;

    for(int i=0; i<w; i++)
        b[i] = read();
    for(int i=w-1; i>0; i--)
        b[i] -= b[i-1];
    for(int i=0; i<w-1; i++)
        b[i] = b[i+1];
    w--;
    if(w==0)//子中长度为1时要特判
        printf("%d\n", n+1);
    else
        printf("%d\n", kmpcnt());
    return 0;
}

#	When	Who	Problem	Lang	Verdict	Time	Memor
41132718	2018-08-02 21:26:50	MoonChasing	D - MUH and Cube Walls	GNU C++	Accepted	31 ms	2400 KB
41132612	2018-08-02 21:23:05	MoonChasing	D - MUH and Cube Walls	GNU C++	Wrong answer on test 2	30 ms	2400 KB