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
思路：将第一次出现的i减去自身长度就可以了
代码：
#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
using namespace std;
void kmp_pre(int x[],int m,int next[]) {
int i,j;
j=next[0]=-1;
i=0;
while(i<m) {
while(-1!=j && x[i]!=x[j])j=next[j];
next[++i]=++j;
}
}
int next1[1000010];
int KMP_Count(int x[],int m,int y[],int n) { //x 是模式串，y 是主串
int i,j;
int ans=-1;
//preKMP(x,m,next);
kmp_pre(x,m,next1);
i=j=0;
while(i<n) {
while(-1!=j && y[i]!=x[j])j=next1[j];
i++;
j++;
// cout<<m<<endl;
if(j>=m) {
ans=i+1-m;
// cout<<next[j]<<endl;
j=next1[j];
break;
}
}
return ans;
}
int a[10005];
int b[1000005];
int main() {
int T;
cin>>T;
int lena,lenb;
while(T--) {
scanf("%d %d",&lena,&lenb);
for(int t=0;t<lena;t++)
{
scanf("%d",&a[t]);
}
for(int t=0;t<lenb;t++)
{
scanf("%d",&b[t]);
}
printf("%d\n",KMP_Count(b,lenb,a,lena));
}
return 0;
}
