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
题意就是从母串中找到子串的开始位置,如果能找到就返回这个位置,如果找不到就返回-1;
这个也算是模板题,下面给出代码。
//注意不要用cin 因为cin输入时间比scanf时间长 容易时间超限
#include<cstdio>
#include<iostream>
#include<cstring>
using namespace std;
const int maxn=1e6+10;
int n,m;
int s[maxn]; //母串
int t[maxn]; //子串
int net[maxn];
void getnext(){
int i=0,j=-1;
net[0]=-1;
while(i<m){
if(j==-1 || t[i]==t[j]){
i++;j++;net[i]=j;
}
else j=net[j];
}
}
int kmp(){
int i=0,j=0;
while(i<n&&j<m){
if(j==-1||s[i]==t[j]){
i++;j++;
}
else j=net[j];
}
if(j==m) return i-m+1;
else return -1;
}
int main(){
int T;
scanf("%d",&T);
while(T--){
cin>>n>>m;
for(int i=0;i<n;i++) scanf("%d",&s[i]);
for(int i=0;i<m;i++) scanf("%d",&t[i]);
getnext();
printf("%d\n",kmp());
}
return 0;
}