HDU1711-Number Sequence

Number Sequence

Problem Description

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

#include<cstdio>///HDU1711-Number Sequence
using namespace std;
int a[1000005],b[10005],next[10005];
int n,m;
int getnext()
{
    int j=1,k=0;
    next[1]=0;
    while(j<=m)
    {
        if(k==0||b[j]==b[k])
        {
            j++;k++;
            next[j]=k;
        }
        else
            k=next[k];
    }
}
int kmp()
{
    getnext();
    int i=1,j=1;
    while(i<=n)
    {
        if(j==0||a[i]==b[j])
        {
            i++;
            j++;
        }
        else
            j=next[j];
        if(j==m+1)
        {
            return i-m;
        }
    }
    return -1;
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d %d",&n,&m);
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
        }
        for(int j=1;j<=m;j++)
        {
            scanf("%d",&b[j]);
        }
        printf("%d\n",kmp());
    }
    return 0;
}