# 山东省第八届 ACM 省赛 company （贪心、水）

## Problem Description

There are n kinds of goods in the company, with each of them has a inventory of cnti and direct unit benefit vali. Now you find due to price changes, for any goods sold on day i, if its direct benefit is val, the total benefit would be i⋅val.

Beginning from the first day, you can and must sell only one good per day until you can’t or don’t want to do so. If you are allowed to leave some goods unsold, what’s the max total benefit you can get in the end?

## Input

The first line contains an integers n(1≤n≤1000).

The second line contains n integers val1,val2,..,valn(−100≤vali≤100).

The third line contains n integers cnt1,cnt2,..,cntn(1≤cnti≤100).

## Output

Output an integer in a single line, indicating the max total benefit.

## Example Input

4
-1 -100 5 6
1 1 1 2


## Example Output

51


## 题意

n 种商品以及这些商品的价值与数量，每天只能购买一件商品，且 ans 为 $sum(t_i×val_i)$ ，其中 $t_i$ 为第几天，求最终所能得到的最大 ans

## AC 代码

#include<iostream>
#include<algorithm>
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<iostream>
using namespace std;
#include<queue>
#include<map>
#define eps (1e-8)
typedef long long LL;
const int mod = 1e9+7;
const int maxn = 100000;
int a[maxn],sum[maxn],d[maxn];
int main()
{
int n;
while(~scanf("%d",&n))
{
int top=n;
for(int i=0; i<n; i++)
scanf("%d",a+i);
for(int i=0; i<n; i++)
{
scanf("%d",d+i);
for(int j=1; j<d[i]; j++)
a[top++]=a[i];
}
sort(a,a+top);
sum[0]=a[0];
LL ans=a[0];
for(int i=1; i<top; i++)
{
sum[i]=sum[i-1]+a[i];
ans+=(i+1)*a[i];
}
for(int i=0; i<top; i++)
{
LL s=ans-a[i]-sum[top-1]+sum[i];
if(s>ans)
ans=s;
else break;
}
cout<<ans<<endl;
}
return 0;
}