将 O(n^2) 的复杂度降低到至少 O(n log n)

Question

I want to reduce complexity of this program but I don't know how.

Program:
You are holding an event and you need money. Every person that participates will give percentage of their salary to the event, but every person must give the same percentage. If the percentage you chose is for example 30% and person P is only willing to give 20%, then person P wont give anything. If you chose 30% and person P is willing to give 80%, then person P will only give 30%.

You want to figure out what is the percentage that will get you the most money.

Inputs are:

First line: Number of people that participate (n)
Next n lines: Salary of i-th person, Max percentage of salary that i-th person is willing to give away

Output:
First line: Largest amount of money that you can get for the event.

Example:

Input:

 4 100001 83.2 40001 20 90001 77.32 300001 1.88

Output:

 146909.5464

Program chose that the percentage that will bring the most money is 77.32 and wrote that the money you can get from that percentage is (100001+90001)*77.32

#include <iostream>
#include <iomanip>

using namespace std;

int N,A[200000];
double P[200000],suma,maxsuma;

int main(){

ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);

cin>>N;
for(int i=0;i<N;i++){
    cin>>A[i]>>P[i];
}
for(int i=0;i<N;i++){
    suma=0;
    for(int j=0;j<N;j++){
        if(P[j]>=P[i]){
            suma+=P[i]*A[j]/100.0;
        }
    }
    maxsuma=max(suma,maxsuma);
}

cout<<fixed<<setprecision(20)<<maxsuma;

}

Answer 1

First, sort all your people by the percentage they are willing to give descending. Now whatever percentage you pick, only some first people on the list would actually pay. Therefore, you would make the most profit-making it the last person percentage. Now also notice, that from this prefix everyone gives salary * percentage , so the total payment would be total salary * percentage . If s[i] is the total salary from all people up to i-th index, than s[i] = s[i - 1] + a[i] where a[i] is i-th person salary. Therefore you just write a loop that would first count s[i] using known s[i - 1] from the previous iteration, and then multiply it by the percentage of i-th person; across all such values pick the highest one.

I modified your code to use this idea.

#include <iostream>
#include <iomanip>
#include <algorithm>

using namespace std;

int N;
std::pair<int, int> A[200000]
double maxsuma;

int main(){

ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);

cin>>N;
for(int i=0;i<N;i++){
    cin>>A[i].second>>A[i].first;
}
std::sort(A, A + N, std::greater<std::pair<int, int>>());
int suma = 0;
for(int i=0;i<N;i++){
    suma += A[i].second;
    maxsuma=max(suma * A[i].first / 100.0, maxsuma);
}

cout<<fixed<<setprecision(20)<<maxsuma;

}