C++ algorithm to calculate least common multiple f

Question

Is there a C++ algorithm to calculate the least common multiple for multiple numbers, like lcm(3,6,12) or lcm(5,7,9,12)?

Answer 1

You can use std::accumulate and some helper functions:

#include <iostream>
#include <numeric>

int gcd(int a, int b)
{
    for (;;)
    {
        if (a == 0) return b;
        b %= a;
        if (b == 0) return a;
        a %= b;
    }
}

int lcm(int a, int b)
{
    int temp = gcd(a, b);

    return temp ? (a / temp * b) : 0;
}

int main()
{
    int arr[] = { 5, 7, 9, 12 };

    int result = std::accumulate(arr, arr + 4, 1, lcm);

    std::cout << result << '\n';
}

Answer 2

/*

Copyright (c) 2011, Louis-Philippe Lessard
All rights reserved.

Redistribution and use in source and binary forms, with or without modification, are permitted provided that the following conditions are met:

Redistributions of source code must retain the above copyright notice, this list of conditions and the following disclaimer.
Redistributions in binary form must reproduce the above copyright notice, this list of conditions and the following disclaimer in the documentation and/or other materials provided with the distribution.
THIS SOFTWARE IS PROVIDED BY THE COPYRIGHT HOLDERS AND CONTRIBUTORS "AS IS" AND ANY EXPRESS OR IMPLIED WARRANTIES, INCLUDING, BUT NOT LIMITED TO, THE IMPLIED WARRANTIES OF MERCHANTABILITY AND FITNESS FOR A PARTICULAR PURPOSE ARE DISCLAIMED. IN NO EVENT SHALL THE COPYRIGHT HOLDER OR CONTRIBUTORS BE LIABLE FOR ANY DIRECT, INDIRECT, INCIDENTAL, SPECIAL, EXEMPLARY, OR CONSEQUENTIAL DAMAGES (INCLUDING, BUT NOT LIMITED TO, PROCUREMENT OF SUBSTITUTE GOODS OR SERVICES; LOSS OF USE, DATA, OR PROFITS; OR BUSINESS INTERRUPTION) HOWEVER CAUSED AND ON ANY THEORY OF LIABILITY, WHETHER IN CONTRACT, STRICT LIABILITY, OR TORT (INCLUDING NEGLIGENCE OR OTHERWISE) ARISING IN ANY WAY OUT OF THE USE OF THIS SOFTWARE, EVEN IF ADVISED OF THE POSSIBILITY OF SUCH DAMAGE.

*/

unsigned gcd ( unsigned a, unsigned b );
unsigned gcd_arr(unsigned * n, unsigned size);
unsigned lcm(unsigned a, unsigned b);
unsigned lcm_arr(unsigned * n, unsigned size);
int main()
{
    unsigned test1[] = {8, 9, 12, 13, 39, 7, 16, 24, 26, 15};
    unsigned test2[] = {2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048};
    unsigned result;

    result = gcd_arr(test1, sizeof(test1) / sizeof(test1[0]));
    result = gcd_arr(test2, sizeof(test2) / sizeof(test2[0]));
    result = lcm_arr(test1, sizeof(test1) / sizeof(test1[0]));
    result = lcm_arr(test2, sizeof(test2) / sizeof(test2[0]));

    return result;
}


/**
* Find the greatest common divisor of 2 numbers
* See http://en.wikipedia.org/wiki/Greatest_common_divisor
*
* @param[in] a First number
* @param[in] b Second number
* @return greatest common divisor
*/
unsigned gcd ( unsigned a, unsigned b )
{
    unsigned c;
    while ( a != 0 )
    {
        c = a;
        a = b%a;
        b = c;
    }
    return b;
}

/**
* Find the least common multiple of 2 numbers
* See http://en.wikipedia.org/wiki/Least_common_multiple
*
* @param[in] a First number
* @param[in] b Second number
* @return least common multiple
*/
unsigned lcm(unsigned a, unsigned b)
{
    return (b / gcd(a, b) ) * a;
}

/**
* Find the greatest common divisor of an array of numbers
* See http://en.wikipedia.org/wiki/Greatest_common_divisor
*
* @param[in] n Pointer to an array of number
* @param[in] size Size of the array
* @return greatest common divisor
*/
unsigned gcd_arr(unsigned * n, unsigned size)
{
    unsigned last_gcd, i;
    if(size < 2) return 0;

    last_gcd = gcd(n[0], n[1]);

    for(i=2; i < size; i++)
    {
        last_gcd = gcd(last_gcd, n[i]);
    }

    return last_gcd;
}

/**
* Find the least common multiple of an array of numbers
* See http://en.wikipedia.org/wiki/Least_common_multiple
*
* @param[in] n Pointer to an array of number
* @param[in] size Size of the array
* @return least common multiple
*/
unsigned lcm_arr(unsigned * n, unsigned size)
{
    unsigned last_lcm, i;

    if(size < 2) return 0;

    last_lcm = lcm(n[0], n[1]);

    for(i=2; i < size; i++)
    {
        last_lcm = lcm(last_lcm, n[i]);
    }

    return last_lcm;
}

Source code reference

Answer 3

Using the fact that lcm should be divisible by all the numbers in list. Here the list is a vector containing numbers

        int lcm=*(len.begin());
    int ini=lcm;
    int val;
    int i=1;
    for(it=len.begin()+1;it!=len.end();it++)
    {
        val=*it;
        while(lcm%(val)!=0)
        {
            lcm+=ini;
        }
        ini=lcm;
    }
    printf("%llu\n",lcm);
    len.clear();

Answer 4

#include
#include

void main()
{
    clrscr();

    int x,y,gcd=1;

    cout<>x;

    cout<>y;

    for(int i=1;i<1000;++i)
    {
        if((x%i==0)&&(y%i==0))
        gcd=i;
    }

    cout<<"\n\n\nGCD :"<
    cout<<"\n\n\nLCM :"<<(x*y)/gcd;

    getch();
}

Answer 5

Using GCC with C++14 following code worked for me:

#include <algorithm>
#include <vector>

std::vector<int> v{4, 6, 10};    
auto lcm = std::accumulate(v.begin(), v.end(), 1, [](auto & a, auto & b) {
    return abs(a * b) / std::__gcd(a, b);
});

In C++17 there is std::lcm function (http://en.cppreference.com/w/cpp/numeric/lcm) that could be used in accumulate directly.

Answer 6

• let the set of numbers whose lcm you wish to calculate be theta
• let i, the multiplier, be = 1
• let x = the largest number in theta
• x * i
• if for every element j in theta, (x*i)%j=0 then x*i is the least LCM
• if not, loop, and increment i by 1