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First N natural can be divided into two sets with given difference and co-prime sums in C++
In this tutorial, we have to find whether the natural numbers from 1 to n is divided into two halves or not. It has to satisfy the following conditions.
The absolute difference between the two series sum should be m.
And the GCD of two sums should be 1 i.e.., co-primes.
The sum of first n natural numbers is (n*(n+1))/2. We can find the sumOne and sumTwo as we have total sum and difference m. See the below equations.
sumOne + sumTwo = (n*(n+1))/2 sumOne - sumTwo = m
Example
Check whether the absolute sum is equal to m or not. And then check for GCD.
#include <bits/stdc++.h>
using namespace std;
bool canSplitIntoTwoHalves(int n, int m) {
int total_sum = (n * (n + 1)) / 2;
int sumOne = (total_sum + m) / 2;
int sumTwo = total_sum - sumOne;
if (total_sum < m) {
return false;
}
if (sumOne + sumTwo == total_sum &&sumOne - sumTwo == m) {
return (__gcd(sumOne, sumTwo) == 1);
}
return false;
}
int main() {
int n = 10, m = 17;
if (canSplitIntoTwoHalves(n, m)) {
cout << "Can split";
}
else {
cout << "Can't split";
}
return 0;
}Output
If you run the above code, then you will get the following result.
Can split
Conclusion
If you have any queries in the tutorial, mention them in the comment section.
Updated on: 29-Dec-2020
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