# Number Sequence

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 40388    Accepted Submission(s): 16659

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

## AC代碼

``````#include <bits/stdc++.h>
#define ll long long
#define ms(a) memset(a,0,sizeof(a))
using namespace std;
const int maxn=1e6+10;
int a[maxn],b[maxn];
int Next[maxn];
int  n,m;
// Next數組下標從1開始
void getNext()
{
int j,k;
j=0;k=-1;Next[0]=-1;
while(j<m)
{
if(k==-1||b[j]==b[k])
Next[++j]=++k;
else
k=Next[k];
}
}
/*

*/
int KMP_Index()
{
int i=0,j=0;
getNext();
while(i<n&&j<m)
{
if(j==-1||a[i]==b[j])
{
i++;j++;
}
else
j=Next[j];
}
if(j==m)
return i-m;
else
return -1;
}
int main(int argc, char const *argv[])
{
int t;
scanf("%d",&t);
while(t--)
{
scanf("%d%d",&n,&m);
ms(a);
ms(b);
ms(Next);
for(int i=0;i<n;i++)
scanf("%d",&a[i]);
for(int i=0;i<m;i++)
scanf("%d",&b[i]);
if(KMP_Index()!=-1)
printf("%d\n",KMP_Index()+1);
else
printf("%d\n",KMP_Index());
}
return 0;
}``````