Number Sequence （KMP第一次出现位置）

Given two sequences of numbers : a[1], a[2], ...... , a[N], and b[1], b[2], ...... , b[M] (1 <= M <= 10000, 1 <= N <= 1000000). Your task is to find a number K which make a[K] = b[1], a[K + 1] = b[2], ...... , a[K + M - 1] = b[M]. If there are more than one K exist, output the smallest one. 

Input

The first line of input is a number T which indicate the number of cases. Each case contains three lines. The first line is two numbers N and M (1 <= M <= 10000, 1 <= N <= 1000000). The second line contains N integers which indicate a[1], a[2], ...... , a[N]. The third line contains M integers which indicate b[1], b[2], ...... , b[M]. All integers are in the range of [-1000000, 1000000]. 

Output

For each test case, you should output one line which only contain K described above. If no such K exists, output -1 instead. 

Sample Input

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

Sample Output

6
-1

思路：将第一次出现的i减去自身长度就可以了

代码：

#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>

using namespace std;
void kmp_pre(int x[],int m,int next[]) {
	int i,j;
	j=next[0]=-1;
	i=0;
	while(i<m) {
		while(-1!=j && x[i]!=x[j])j=next[j];
		next[++i]=++j;
	}
}
int next1[1000010];
int KMP_Count(int x[],int m,int y[],int n) { //x 是模式串，y 是主串
	int i,j;
	int ans=-1;
//preKMP(x,m,next);
	kmp_pre(x,m,next1);
	i=j=0;
	while(i<n) {
		while(-1!=j && y[i]!=x[j])j=next1[j];
		i++;
		j++;
//		cout<<m<<endl;
		if(j>=m) {
			ans=i+1-m;
//		    cout<<next[j]<<endl;
			j=next1[j];
			break;
		
		}
	}
	return ans;
}

int  a[10005];
int b[1000005];

int main() {
	int T;
	cin>>T;
	int lena,lenb;
	while(T--) {
		scanf("%d %d",&lena,&lenb);
	    for(int t=0;t<lena;t++)
	    {
	    	scanf("%d",&a[t]);
		}
		for(int t=0;t<lenb;t++)
		{
			scanf("%d",&b[t]);
		}
		printf("%d
",KMP_Count(b,lenb,a,lena));
	}


	return 0;
}