ACM 第八届山东省赛 J company SDUT 3902

最新推荐文章于 2018-04-14 15:31:00 发布

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本文介绍了一种基于贪心算法的编程题解决方案，通过倒序处理商品库存与收益，实现利益最大化。问题背景为一家公司拥有多种商品，在价格波动下如何安排销售顺序以获得最大总收益。

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company

Time Limit: 1000MS Memory Limit: 65536KB

Submit Statistic

Problem Description

There are n kinds of goods in the company, with each of them has a inventory of and direct unit benefit . 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≤.≤100).
The third line contains n integers cnt1,cnt2,..,cntn(1≤≤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

Hint

sell goods whose price with order as -1, 5, 6, 6, the total benefit would be -1*1 + 5*2 + 6*3 + 6*4 = 51.

Author

“浪潮杯”山东省第八届ACM大学生程序设计竞赛（感谢青岛科技大学）

贪心算法 ,只能用一个for循环并且用long long int 数据量会非常大

倒着从后面考虑叠楼梯

层数代表天数有前到后依次增加

代码:

#include <iostream>
#include <algorithm>
#include <cmath>
#include <cstring>
#include <stdio.h>
#include <string>

typedef long long ll;
using namespace std;
ll a[1000000];
ll dp[1000000];
int main()
{
    int k,i,j,n;
    while(cin>>n)
    {
        memset(a,0,sizeof(a));
        memset(dp,0,sizeof(dp));
        for(i=0;i<n;i++)
        {
            cin>>a[i];
        }
        for(i=0,j=0;i<n;i++)
        {
            cin>>k;
            while(k>1)
            {
                a[n+j]=a[i];
                j++;
                k--;
            }
        }

        sort(a,a+n+j);
        int len=n+j;
        ll ans=0;
        ll dsum=0;
        //dp[len-1]=a[len-1];
        for(i=len-1;i>=0;i--)
        {
            dsum+=a[i];
            if(dsum<0)
                break;
            dp[i]=dp[i+1]+dsum;
            if(ans<dp[i])
                ans=dp[i];
        }
        cout<<ans<<endl;
    }
    return 0;
}
/*
5
-1 -1 -1 -1 1
1 1 1 1 1

*/