ZCCZCC Loves Codefires

ZCC Loves Codefires

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 0    Accepted Submission(s): 0

Problem Description

   Though ZCC has many Fans, ZCC himself is a crazy Fan of a coder, called "Memset137".   It was on Codefires(CF), an online competitive programming site, that ZCC knew Memset137, and 
immediately became his fan.   But why?   Because Memset137 can solve all problem in rounds, without unsuccessful submissions; his estimation of 
time to solve  certain problem is so accurate, that he can surely get an Accepted the second he has predicted. 
He soon became IGM, the  best title of Codefires. Besides, he is famous for his coding speed and the 
achievement in the field of Data Structures.   After become IGM, Memset137 has a new goal: He wants his score in CF rounds to be as large as possible.   What is score? In Codefires, every problem has 2 attributes, let's call them Ki and Bi(Ki, Bi＞0). if
 Memset137 solves the
  problem at Ti-th second, he gained Bi-Ki*Ti score. It's guaranteed Bi-Ki*Ti is always positive during the round time.   Now that Memset137 can solve every problem, in this problem, Bi is of no concern. Please write a program 
to calculate the  minimal score he will lose.(that is, the sum of Ki*Ti).

 

Input

   The first line contains an integer N(1≤N≤10^5), the number of problem in the round.   The second line contains N integers Ei(1≤Ei≤10^4), the time(second) to solve the i-th problem.   The last line contains N integers Ki(1≤Ki≤10^4), as was described.

 

Output

   One integer L, the minimal score he will lose.

 

Sample Input

310 10 201 2 3

 

Sample Output

150
题意：每一个题目有两个权值，要求计算在使用时间之内，找出使用时间和得分最大，计算出时间/题的值，存着，从大到小排序，再将题从大到小排序，依次计算和。

代码：

#include <iostream>
#include <cstdio>
#include <algorithm>
#define N 100005
using namespace std;


struct node
{
    int k,e;
    double f;
};
bool cmp(node a,node b)
{
    return a.f<b.f;
}
node a[N];
int main()
{
    __int64 sum=0,ans=0;
    int n;
    while(scanf("%d",&n)==1)
    {
        for(int i=0;i<n;i++)
        {
            scanf("%d",&a[i].e);
        }
        for(int i=0;i<n;i++)
        {
            scanf("%d",&a[i].k);
            a[i].f=1.0*a[i].e/a[i].k;
        }
        sort(a,a+n,cmp);
        for(int i=0;i<n;i++)
        {
            ans+=(sum+a[i].e)*a[i].k;
            sum+=a[i].e;
        }
        printf("%I64d\n",ans);
    }
    return 0;
}