J--company

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company

Time Limit: 1000MS Memory Limit: 65536KB

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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

51
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.

思路：用数组存起来，将其从小到大排序，若只考虑正数的话只需要 a*i+b*（i+1）+……若加一个负数的话，则正数部分变成了a*(i+1)+b*（i+2）+…… 与原来相比多了 a+b+…… 同时还多了一个负数s

若 a+b +…… >这个负数s的时候，那么这个负数就要计算在内————进一步抽象只需要求一下从当前这个负数到数组最后的和是否满足大于等于0，就好了。找到临界部分就可以了。

提示：

1. 数据量虽然是1000 但是下面的个数可以是100 相当于1000*100 数组要开的足够大。

2. 求和需要 longlongint

#include <stdio.h>#include <string.h>#include <algorithm>using namespace std;bool cmp(int m, int n){    return m<n;}int main (){    int a[150000];    int b[150000];    int n,i,k,j;    long long  int sum;    scanf("%d",&n);    memset(a,0,sizeof(a));    memset(b,0,sizeof(b));    for(i=0; i<n; i++)    {        scanf("%d",&a[i]);    }    for(i=0; i<n; i++)    {        scanf("%d",&b[i]);    }    j=n;    for(i=0; i<n; i++)    {        if(b[i]>1)        {            for(k=1; k<b[i]; k++)            {                a[j]=a[i];                j++;            }        }    }    n=j;    sort(a,a+n,cmp);    if(a[n-1] <=0)    {        printf("0\n");    }    else    {        k=0;        j=0;        sum=0;        for(i=n-1; i>=0; i--)        {                j=j+a[i];                if(j<=0)                {                    k=i+1;                    break;                }        }        int ans=1;        for(i=k;i<n;i++)        {            sum+=a[i]*ans;            ans++;        }        printf("%lld\n",sum);    }    return 0;}

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