题意:
解法:
注意到i必须满足<=j,
因此c[k]=max(c[k-1],max{
a[i]}*b[k]),其中i<=k.
max{
a[i]}开一个变量维护一个前缀max即可.
code:
#include <bits/stdc++.h>
using namespace std;
#define int long long
const int maxm=2e5+5;
int ans[maxm];
int a[maxm];
int b[maxm];
int n;
signed main(){
ios::sync_with_stdio(0);
cin>>n;
for(int i=1;i<=n;i++)cin>>a[i];
for(int i=1;i<=n;i++)cin>>b[i];
int aa=-1;
for(int i=1;i<=n;i++){
aa=max(aa,a[i]);
ans[i]=max(ans[i-1],aa*b[i]);
}
for(int i=1;i<=n;i++){
cout<<ans[i]<<endl;
}
return 0;
}