HDU 1711 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

简单kmp模板题
#include<cstdio>
#include<iostream>
#include<vector>
#include<algorithm>
#include<string>
#include<cstring>
using namespace std;
const int maxn = 1000005;
int nt[maxn],a[maxn],b[maxn],T,n,m,tot;

int main()
{
	scanf("%d",&T);
    while (T--)
    {
        scanf("%d%d",&n,&m);
		for (int i=0;i<n;i++) scanf("%d",&a[i]);
		for (int i=0;i<m;i++) scanf("%d",&b[i]);

        nt[0] = tot=-1;
        for (int i = 0; i<m; i++)
        {
            int k = nt[i];
            while (k >= 0 && b[i] != b[k]) k = nt[k];
            nt[i + 1] = k + 1;
        }
        for (int i = 0, j = nt[1]; i<n;)
        {
            if (j < 0 || a[i] == b[j])
            {
                i++, j++;
				if (j==m) {tot=i-j+1;break;}
            }
            else j = nt[j];
        }
        printf("%d\n", tot);
    }
    return 0;
}