Count Distinct Non-Negative Integer Pairs (x, y) that Satisfy the Inequality x*x + y*y < n in C++

We are given a positive integer N. The goal is to count the pairs of distinct non-negative positive integers that satisfy the inequality − x*x + y*y < N.

We will start from x=0 to x² lt; N and y=0 to y² < N . If any x² + y² < N, increase count of pairs.

Let us understand with examples −

Input − n=4

Output − distinct pairs= 4

Explanation − pairs will be (0,0), (1,1), (0,1), (1,0). All these satisfy the inequality x² + y² < 4

Input −n=2

Output − distinct pairs= 3

Explanation &minus pairs will be (0,0), (0,1), (1,0). All these satisfy the inequality x² + y² < 2

Approach used in the below program is as follows

• The integer N stores the positive integer.

• Function countPairs(int n) takes n as input and returns the count of distinct non-negative positive integer pairs that satisfy the inequality −x² + y² < n.

• count stores the number of such pairs , initially 0.

• Start from i=0 to i2 < n and other loop j=0 to j2 < n.

• If i2 + j2 < n increment count.

• Return count in the end as result.

Example

 Live Demo

#include <iostream>
using namespace std;
int countPairs(int n){
   int count = 0;
   for (int i = 0; i*i < n; i++)
      for (int j = 0; j*j < n; j++) //x*x + y*y < n
      if(i*i + j*j < n)
         count++;
   return count;
}
int main(){
   int N=4;
   cout << "Distinct Non-Negative integer pairs count: " << countPairs(N) ;
   return 0;
}

Output

If we run the above code it will generate the following output −

Distinct Non-Negative integer pairs count: 4

Sunidhi Bansal

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Updated on: 13-Aug-2020

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