A - Number Sequence 哈希算法（例题）

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. 

InputThe 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]. 
OutputFor 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
哈希算法的模板题，题目数据没有当 n < m 的情况

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
const int N=1E6+100;
const int M=1E4+7;
const int base =233;
const int mod=1e9+7;
typedef unsigned long long ll;
ll a[N],b[M];
ll ha[N],p[N];

ll gethash(ll x,ll y){
    return (ha[y]%mod-(ha[x-1]%mod*p[y-x+1]%mod)%mod+mod)%mod;
}

int main(){
    int t;
//    cin>>t;
    scanf("%d",&t);
    while(t--){
        memset(a,0,sizeof(a));
        memset(b,0,sizeof(b));
        ll n,m;
        scanf("%lld%lld",&n,&m); 
        for(int i=1;i<=n;i++)
            scanf("%lld",&a[i]); 
        for(int i=1;i<=m;i++)
            scanf("%lld",&b[i]); 
        p[0]=1;
        if(n>=m){ 
            for(int i=1;i<=n;i++){
                ha[i]=((ha[i-1]%mod*base)%mod+a[i])%mod;
                p[i]=(p[i-1]%mod*base)%mod; 
            }
            ll ans=0;
            for(int i=1;i<=m;i++) ans=(ans%mod*base+b[i])%mod;
            int pos=0;
            for(int i=m;i<=n;i++){
                if(gethash(i-m+1,i)==ans){
                    pos=i-m+1;
                    break;
                }
            }
            if(pos==0) pos=-1;
            printf("%d\n",pos);
        }
//        else {
//            for(int i=1;i<=m;i++){
//                ha[i]=((ha[i-1]%mod*base)%mod+b[i])%mod;
//                p[i]=(p[i-1]%mod*base)%mod; 
//            } 
//            ll ans=0;
//            for(int i=1;i<=n;i++) ans=(ans%mod*base+a[i])%mod;
//            int pos=0;
//            for(int i=n;i<=m;i++){
//                if(gethash(i-n+1,i)==ans){
//                    pos=i-n+1;
//                    break;
//                }
//            }
//            if(pos==0) pos=-1;
//            printf("%d\n",pos);
//        } 
    }
    return 0;
}