2分

时间复杂度O(m longn)

#include<bits/stdc++.h>
#define ll long long
using namespace std;
ll n,m,hi[1145147],ne[1145147],b;
ll rf(ll x){
    ll l=1,r=n,mid,ans;
    while(l<=r){
        mid=(l+r)/2;
        if(ne[mid]>x){
            ans=mid-1;
            r=mid-1;
        }else l=mid+1;
    }
    return hi[ans];
}
int main(){
    cin>>n>>m;
    for(int i=1;i<=n;i++){
        cin>>hi[i];
        ne[i]=max(hi[i]-hi[i-1],ne[i-1]);
    }
    while(m--){
        cin>>b;
        if(b>=ne[n]){
            cout<<hi[n];
        }
        else cout<<rf(b);
        cout<<" ";
    }
    return 0;
}