Member Avatar for arupa

write a program to calculate the Highest Common Factor of three input numbers

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  • Latest Post Latest Post by puneetkay

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For computing the HCF of three integers, the fast, robust choice is the Euclidean approach that pointed to; the brute-force loop suggested by is both slower and (as written) has a loop-bound error — if you try a divisor loop, iterate downward from the smaller number, not upward from M to M/2. For real code, prefer standard library helpers where available (they handle signs and zeros cleanly) or fold a tested gcd routine over the inputs.

Example using Python's standard library (no manual algorithm required):

from math import gcd
from functools import reduce

def hcf_of_three(a, b, c):
    vals = [abs(int(x)) for x in (a, b, c)]
    if all(v == 0 for v in vals):
        return 0  # behaviour: many libraries return 0 for (0,0,0)
    return reduce(gcd, vals)

print(hcf_of_three(100, 125, 75))  # 25

Example using modern C++ (C++17 or later):

#include <numeric>
#include <iostream>
#include <cstdlib>

int main() {
    long long a, b, c;
    if (!(std::cin >> a >> b >> c)) return 0;
    a = llabs(a); b = llabs(b); c = llabs(c);
    if (a==0 && b==0 && c==0) { std::cout << 0 << '\n'; return 0; }
    std::cout << std::gcd(a, std::gcd(b, c)) << '\n';
}

Notes and pitfalls:

  • Take absolute values first so negative inputs behave as expected.
  • Decide how to treat (0,0,0) — many libraries return 0, though mathematically gcd(0,0) is undefined.
  • For C (no std::gcd), either implement an iterative gcd carefully or call a library routine; avoid naive divisor searches for large inputs.
  • Euclid-based gcd runs in about O(log min(a,b)), so it scales well for big integers.

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Member Avatar for selvaganapathy

I give algorithm you try it in C#

HCF (GCD) Algorithm

GCD ( M, N ) {
If N > M then return GCD ( N, M )
else If N = 0 then return M
else return GCD ( N, MOD ( M, N) )
}

It is simple to covert it to C#
Also it is for two input

For three input

GCD ( M, N, O ){
return GCD ( M, GCD ( N, O ) )
}

Member Avatar for puneetkay

I give algorithm you try it in C#

HCF (GCD) Algorithm

GCD ( M, N ) {
If N > M then return GCD ( N, M )
else If N = 0 then return M
else return GCD ( N, MOD ( M, N) )
}

It is simple to covert it to C#
Also it is for two input

For three input

GCD ( M, N, O ){
return GCD ( M, GCD ( N, O ) )
}

Hie,

Thanks buddy, but i didnt really get the algo.
Heres the one from me:
=======
1 Input: num type variables M,N, HCF. (eg: M=100, N=125)
2 Find the smaller one from M,n. (eg: smaller: M)
3 Start a loop from M to M/2. (eg: i=m;i<=m/2;i++)
___1 Check whether i%m=0 & i%n==0.
___2 HCF = i; break;

Hope this can help you! :)

Regards,
PuneetKay

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