Greedy-One-more.cpp

// Ashima has brought home n cats. Now, being a cat lover, she is taking care of the cats and has asked me to bring cat food for them. Being a guy with no idea what to buy, I brought some n packets of cat food (I atleast knew that each and every cat being a good junkie will completely eat a whole packet of cat food and won't share anything with other cats). Each food packet has some calorie value c. If a cat with original strength s eats that packet, the strength of the cat becomes c*s. Now, Ashima is angry at me that I did not know this fact and now all the cats won't be able to eat the maximum strength packet and increase their strength (and annoying powers).

// To calm her mood, I need your help. I will provide you with the original strength of each cat and the calorie value of each of the n packets. Help me by telling me what is the maximum value of sum of the final strengths of the cats that can be obtained if each cat is given a whole packet of cat food to eat.

// Input
// The first line of the input will consist of n, the number of cats as well as the number of food packets brought by me.
// The second line will consist of n space separated integers si, the original strength of the cats.
// Third line consists of n space separated integers ci, the calorie value of food packets.

// Output:
// An integer which is the maximum value of sum of the final strengths of the cats that can be obtained.

// Constraints:
// 1 ≤ n ≤ 106
// 1 ≤ si ≤ 106
// 1 ≤ ci ≤ 106
// SAMPLE INPUT

// 2
// 3 1
// 4 3

// SAMPLE OUTPUT

// 15
//  Explanation

// The maximum sum is obtained by giving packet with calorie value 4 to the first cat and the packet with calorie value 3 to the second cat.


#include<bits/stdc++.h>
using namespace std;
int main()
{  long long int n,i,j;
    cin>>n;
   long long  int k[n+1], c[n+1];
    long long int sum=0;
    for(i=0; i<n; i++)
    {
        cin>>k[i];
    }
    sort(k,k+n);
    for(j=0; j<n; j++)
    {
        cin>>c[j];
    }
    sort(c,c+n);
    for(i=0; i<n; i++)
    {
        sum=sum+k[i]*c[i];
    }
    cout<<sum;
    return 0;
}