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Here we will see how to count the number of pairs of coprime numbers in a range where one number does not appear in more than one pair.
Before discussing logic, let us look at what are coprime numbers? Relatively prime numbers are those numbers that have only one positive integer divisor (i.e. 1). In other words, we can say that the greatest common divisor of these two numbers is 1.
Here we provide lower and upper bounds. If the lower and upper bounds are 1 and 6 respectively, then there are three logarithms. They are (1, 2), (3, 4) and (5, 6)
The way to solve this problem is: if these numbers are continuous, they are a pair of coprime numbers.
are always mutually prime. So the count will be (R – L 1)/2. If (R – L 1) is odd, then 1 The remaining numbers will not be put into any pairs, if they are even, then all will become pairsBegin return (R – L + 1)/2 End
#include <iostream> using namespace std; int countCoPrimePairs(int L, int R) { return (R - L + 1)/2; } main() { int l = 1, r = 6; cout << "Number of co-prime pairs: " << countCoPrimePairs(l, r); }
Number of co-prime pairs: 3
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