D. MUH and Cube Walls(好裸的KMP啊)

题目链接：D. MUH and Cube Walls

time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard 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”.

Examples
inputCopy
13 5
2 4 5 5 4 3 2 2 2 3 3 2 1
3 4 4 3 2
outputCopy
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.

总结：这道题好惨啊，一年前的错思路一直影响到现在~~~~

思路：其实这道题是很简单的KMP题。因为相邻两个数值的差值是固定的，所以只要KMP他们的差值就可以了。
KMP不会的可以看一下：传送门

代码：

#include<stdio.h>
#include<string.h>

const int maxn=2*1e5+9;

int nex[maxn];
int a[maxn],b[maxn];

void KMP(int n)
{
    int i=0,j=-1;
    nex[0]=-1;
    while(i<n)
    {
        if(j==-1||b[i]==b[j])
        {
            i++,j++;
            nex[i]=j;
        }
        else j=nex[j];
    }
}
int main()
{
    int n,m;
    while(~scanf("%d%d",&n,&m))
    {
        for(int i=0;i<n;i++)
        {
            scanf("%d",&a[i]);
            if(i-1>=0)
                a[i-1]=a[i]-a[i-1];
        }
        for(int i=0;i<m;i++)
        {
            scanf("%d",&b[i]);
            if(i>=1)
                b[i-1]=b[i]-b[i-1];
        }

        KMP(m-1);
        int ans=0;
        int i=0,j=0;
        while(i<n-1)//KMP匹配
        {
            if(j==-1||a[i]==b[j])
            {
                i++,j++;
                if(j==m-1)
                {
                    ans++;
                    j=nex[j];
                }
            }
            else j=nex[j];
        }
        if(m==1)//特判一下m==1的情况
            ans++;
        printf("%d\n",ans);
    }
    return 0;
}