Number Sequence-优快云博客

Number Sequence

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 9297 Accepted Submission(s): 4266

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

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#include<iostream>
#include<cstdio>
#include<cstring>
#define M 1000040
using namespace std;
int a[M];
int b[M];
int next[M];
int n,m;
void getnext()
{
    int k,j;
    j=0;
    k=-1;
    next[0]=-1;
    while(j!=m)
    {
      if((k==-1)||(b[j]==b[k]))
      {
          k++;
          j++;
          if(b[j]!=b[k])
          {
             next[j]=k;
          }
          else
          next[j]=next[k];
      }
      else
      k=next[k];
    }
}
int kmp()
{
    int i,j;
    i=0;
    j=0;
    while((i!=n)&&(j!=m))
    {
        if(a[i]==b[j])
        {
            if(j==m-1)
            {
                return i-j+1;
            }
            ++i;
            ++j;
        }
        else
        {
            if(next[j]!=-1)
            {
                j=next[j];
            }
            else
            {
                ++i;
                j=0;

            }
        }
    }
    if(j!=m-1)
    return -1;
}
int main()
{
    int i,j,k,t;
    scanf("%d",&t);
    while(t--)
    {
    scanf("%d%d",&n,&m);
    for(i=0;i<n;i++)
    scanf("%d",&a[i]);
    for(i=0;i<m;i++)
    scanf("%d",&b[i]);
    getnext();
    k=kmp();
    printf("%d\n",k);
    }
    return 0;
}