山东省第八届acm大赛 J题 （SDUT 3902）

company

Time Limit: 1000MS Memory Limit: 65536KB

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 61 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.

WA了无数次   最后结果记住要用 long long    

AC代码：

#include <stdio.h>  #include <algorithm>  using namespace std;    int cmp(int a,int b){      return a>b;  }  int main (){      int num[1005];      int a[100010];      int n;      scanf ("%d",&n);      for (int i=1;i<=n;i++){          scanf ("%d",&num[i]);      }      int len=0;      int v;      for (int i=1;i<=n;i++){          scanf ("%d",&v);          while (v--){              a[len++]=num[i];          }      }        sort(a,a+len,cmp);    //printf("%d\n",len);     long long maxn=-999999999;      int flag=1;      for (int k=len;k>0&&flag;k--)//    要有flag标志位   不然会超时      {          flag=0;          long long sum=0;          for (int i=0;i<=k;i++){              sum+=(a[i]*(k-i));          }          if (maxn<sum){              flag=1;              maxn=sum;          }      }        printf ("%lld\n",maxn);      return 0;  }