Question

I have to build a program as bellow:

User enters 3 Integers: a, b, c
User enters one integer n, and then n integers (ascending, and different numbers)
Program checks all possible pairs (x,y) (with x!=y) of the numbers entered if verifies the equation ax^2 + by^2 = c and prints all pairs which verifies the equation.

I done the program as bellow:

#include<iostream>
using namespace std;
int main() {
int a,b,c,n,i,j;
cin >> a;
cin >> b;
cin >> c;
cin >> n;
int num[n];
for(i=0;i<n;i++) {
   cin >> num[i];
}
for (i=0;i<n;i++)
for (j=i+1;j<n;j++) {
    if(a*num[i]*num[i]+b*num[j]*num[j] == c) {
      cout << "(" << num[i] << "," << num[j] << ")";
    }
    if(a*num[j]*num[j]+b*num[i]*num[i] == c) {
      cout << "(" << num[j] << "," << num[i] << ")";
    }
 }
 return 0;
 }

I made it by O(nlogn) with two 'for' statements but i know it could be done by O(n).

NOTE THAT MY PROGRAM WORKS AND I DON'T NEED TO ADD EXPECTED OUTPUT AND MY CURRENT OUTPUT AS YOU SAID IN THE COMMENTS. I ONLY WANT IT TO BE O(N) not O(nlogn) -> I WANT AN OPTIMIZED VERSION OF THE CODE!

How can I do this?

Example of running program: a=1, b=1, c=20 Then n = 5 Then n numbers: 2 3 4 9 18 Program will show all pairs (x,y) which verifies the equation x^2 + y^2 = 20. In this case it shows (2,4) and (4,2).

Thank you!

Answer 1

Assuming 0 based index...

Set i=0
Set j=n-1
While i<n or j>=0
    Set sum=a(num[i]^2)+b(num[j^2)
    If sum==c then found pair, and increase i
    If sum<c increase i
    If sum>c decrease j

Answer 2

我发现这个问题在这里解决了：http://lonews.ro/educatie-cultura/22899-rezolvare-admitere-universitatea-bucuresti-2015-pregatire-informatica.html并稍微改了一下以显示对（它最初显示了对的数量）。

#include <iostream>
using namespace std;

int main()
{
   int a, b, c, n, i=0;
   cout << "a = ";
   cin >> a;
   cout << "b = ";
   cin >> b;
   cout << "c = ";
   cin >> c;
   cout << "n = ";
   cin >> n;

   int s[n];
   for(i=0; i<n; i++) {
       cout << "s[" << i+1 << "] = ";
       cin >> s[i];
   }
   int j=n-1;
   i = 0;
   while(j>=0 || i<n) {
       if(a*s[i]*s[i] + b*s[j]*s[j] == c) {
           cout << "(" << s[i] << "," << s[j] << ") ";
           i++;
       }
       if(a*s[i]*s[i] + b*s[j]*s[j] < c) {
           i++;
       }
       if(a*s[i]*s[i] + b*s[j]*s[j] > c) {
           j--;
       }
   }
   return 0;
}

Pairs x,y verify equation

2 个答案: