ZCC Loves Codefires

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

3
10 10 20
1 2 3

Sample Output

150

Hint
Memset137 takes the first 10 seconds to solve problem B, then solves problem C at the end of the 30th second. Memset137 gets AK at the end of the 40th second.
L = 10 * 2 + (10+20) * 3 + (10+20+10) * 1 = 150. 
 

Author

镇海中学

Source

2014 Multi-University Training Contest 2 

分析：假设一种情况的做完题的先后顺序确定，那么失分也就确定，现在失分要尽可能的小，那么就得交换位置，假设 i 与 j 是相邻的位置，那么交换两个相邻位置就只有一个值在变 Ei*Kj ---->  Ej*Ki，所以要交换得满足条件：Ei*Kj >Ej*Ki   <==> Ei/Ki > Ej/Kj 。

#include<stdio.h>
#include<algorithm>
using namespace std;
struct node
{
    int e,k;
    double f;
};
bool cmp(node a,node b)
{
    return a.f<b.f;
}
node a[100005];
int main()
{
    int n;
    __int64 sumt=0,ans=0;
    scanf("%d",&n);
    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+=(sumt+a[i].e)*a[i].k; sumt+=a[i].e;
    }
    printf("%I64d\n",ans);
}