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
代码如下：
#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
using namespace std;
int n[];
int s[],a[];
int len1,len2; void kmp()
{
n[]=;
for(int i=,k=;i<len2;i++)
{
while(k>&&a[k]!=a[i]) k=n[k-];
if(a[k]==a[i]) k++;
n[i]=k;
}
} int main()
{
///cout << "Hello world!" << endl;
int T;
cin>>T;
while(T--)
{
scanf("%d%d",&len1,&len2);
for(int i=;i<len1;i++)
scanf("%d",&s[i]);
for(int i=;i<len2;i++)
scanf("%d",&a[i]);
kmp();
int flag=-;
for(int i=,j=;i<len1;i++)
{
while(j>&&a[j]!=s[i]) j=n[j-];
if(a[j]==s[i]) j++;
if(j==len2) { flag=i-len2+; break; }
}
if(flag>=) printf("%d\n",flag+);
else puts("-1");
}
return ;
}