A - Number Sequence

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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

这道题将KMP模板中的两个字符串匹配改成了两个数组，原理还是一样的。
本人并不太清楚KMP的原理，只是在模板的基础上进行了修改。

#include <stdio.h>
using namespace std;
const int N = 1e6;
int Next[N + 5];
int s1[N + 5], s2[N + 5];
int len1, len2;

void get_next()
{
    int i, j;
    i = 0;
    Next[0] = j = -1;//第一个next值为-1 
    while(i < len2) {
        if(j == -1 || s2[i] == s2[j]) Next[++i] = ++j;//如果最长前缀和最长后缀相同 
        else j = Next[j];//否则回退
    }
}


void kmp()
{
    int i, j;
    i = j = 0;
    while(i < len1) {
        if(j == -1 || s1[i] == s2[j]) ++i, ++j;
        else j = Next[j];
        if(j == len2){
            printf("%d\n",i - len2 + 1);
            break;
    	}
	}
    if(j!=len2)
    	printf("-1\n");
}


int main()
{
	int t;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d%d",&len1,&len2);
	    for(int i=0;i<len1;i++)
	    	scanf("%d",&s1[i]);
	    for(int i=0;i<len2;i++)
	    	scanf("%d",&s2[i]);
	    get_next();
	    kmp(); 
	}
    return 0;
}