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
题意： 给你一个长度为n的字串 一个长度为m的模板串
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<vector>
#include<queue>
#define ll long long
using namespace std;
int a[1000010],b[1000010];
int nexxt[1000010];
int n,m;
void getnext()
{
int j=-1,i=0;
nexxt[0]=-1;
while(i<m)
{
if(j==-1||b[i]==b[j])
{
i++;
j++;
nexxt[i]=j;
}
else
j=nexxt[j];
}
}
int kmp()
{
getnext();
int j=0,i=0;
while(i<n)
{
if(j==-1||a[i]==b[j])
i++,j++;
else
{
j=nexxt[j];
}
if(j==m)
{
return i-m+1;
}
}
return -1;
}
int main()
{
ios::sync_with_stdio(false);
int t;
cin>>t;
while(t--)
{
cin>>n>>m;
for(int i=0;i<n;i++)
cin>>a[i];
for(int i=0;i<m;i++)
{
cin>>b[i];
}
cout<<kmp()<<endl;
}
return 0;
}
模板：
void getnext()
{
int j=-1,i=0;
nexxt[0]=-1;
while(i<m)
{
if(j==-1||b[i]==b[j])
{
i++;
j++;
nexxt[i]=j;
}
else
j=nexxt[j];
}
}
int kmp()
{
getnext();
int j=0,i=0;
while(i<n)
{
if(j==-1||a[i]==b[j])
i++,j++;
else
{
j=nexxt[j];
}
if(j==m)
{
//返回你所需要的东西
}
}
return ;//返回你所需要的东西
}