hdu1711Number 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

 

Source

HDU 2007-Spring Programming Contest

#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>

using namespace std;

int an[1100000];
int bn[1100000];
int nextt[1100000];
int match[1100000];

int main()
{
    int z,n,m,i,j,k;
    cin>>z;
    while(z--)
    {
        scanf("%d%d",&n,&m);
        for(i = 1;i<=n;i++)
            scanf("%d",&an[i]);
        for(i = 1;i<=m;i++)
            scanf("%d",&bn[i]);
        nextt[1] = 0;
        for(i = 2;i<=m;i++)
        {
            int t = nextt[i-1];
            while(t&&bn[i]!=bn[t+1]) t = nextt[t];
            if(bn[i] == bn[t+1]) t++;
            nextt[i] = t;
        }
        match[0] = 0;
        bool b = 0;
        for(i = 1;i<=n;i++)
        {
            int t = match[i-1];
            while(t&&an[i] != bn[t+1]) t = nextt[t];
            if(an[i] == bn[t+1]) t++;
            match[i] = t;
            if(t == m)
            {
                cout<<i-m+1<<endl;
                b = 1;
                break;
            }
        }
        if(!b) cout<<-1<<endl;
    }
    return 0;
}