company

题目描述

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?

输入

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).

输出

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

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.

样例输入

4-1 -100 5 61 1 1 2

样例输出

51

前缀和的应用

#include <algorithm>#include <iostream>#include <cstring>#include <cstdio>#include <string>using namespace std;#define mod 1000000007#define INF 0x3f3f3f3f#define N 1000int val[N],cnt[N];int b[N*100];int sum[N*100];int main(){    int n;    scanf("%d",&n);    for(int i = 0; i < n; i++)        scanf("%d",&val[i]);    for(int i = 0; i < n; i++)        scanf("%d",&cnt[i]);    int k = 0;    for(int i = 0; i < n; i++) {        for(int j = 0; j < cnt[i]; j++) {            b[k++] = val[i];        }    }    k--;    sort(b,b+k+1);  //  for(int i = 0;i <= k;i++) //       printf("%d ",b[i]); //   cout << endl;    long long ans = 0,t;    for(int i = 0; i <= k; i++) {        if(i == 0) {            sum[i] = b[i];            ans = b[i];            continue;        }        sum[i] = sum[i-1] + b[i];        ans += (i + 1) * b[i];    }    for(int i = 0; i <= k; i++) {        t = ans  - (sum[k] - sum[i] + b[i]);        if(t > ans)            ans = t;    }    printf("%lld\n",ans);    return 0;}