思路

暴力枚举,O(nm) 注意要等长,所以母串最多到n-m

代码

#pragma GCC optimize("Ofast", "inline", "-ffast-math")
#pragma GCC target("avx,sse2,sse3,sse4,mmx")
#include<bits/stdc++.h>
#define inf 0x3f3f3f3f
#define int long long
using namespace std;
const int N=2e5+7;
const int mod=1e9+7;

//int read(){	int x=0,f=1;char ch=getchar();while(ch<'0'||ch>'9'){if(ch=='-') f=f*-1;ch=getchar();}while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}return x*f;}

int n,m,a[N],b[N],ans=inf;

signed main(){
//	ios::sync_with_stdio(0);
//	cin.tie(0);cout.tie(0);
//  freopen("in.cpp","r",stdin);
//  freopen("out.cpp","w",stdout);
	cin>>n;
	for(int i=1;i<=n;i++) cin>>a[i];
	cin>>m;
	for(int i=1;i<=m;i++) cin>>b[i];
	for(int i=0;i<=m-n;i++){
		int tmp=0;
		for(int j=1;j<=n;j++){
			tmp+=(b[i+j]-a[j])*(b[i+j]-a[j]);	
		}
		ans=min(tmp,ans);
	} 
	cout<<ans<<"\n";
	return 0;
}