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 < N and y=0 to y2 < N . If any x2 + y2 < N, increase count of pairs.
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 − 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
Distinct Non-Negative integer pairs count: 4
Published on 03-Aug-2020 10:12:59
