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Algorithm - GCD and LCM problems

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问题 Input for this problem is an array A of positive integers and single positive integer k. The output of the program is True if k is in the set S defined below, False otherwise. Define the set S as follows: if x is in A then x is in S if x and y are in S, then GCD(x,y) is in S if x and y are in S, then LCD(x,y) is in S Additional constraints: The size of the array A is ≤ 50000, k ≤ 10 12 , and x ≤ 10 12 for each x in A. The program must return an answer in 1 second or less. I don't have any

GCD and LCM relation

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问题 The following relation works only for two (3, 12) numbers, it fails to produce the right answer when used for three numbers (3,12,10) . Just wondering if is it my understanding or it is just for two numbers and for me same is true for Euclid algorithm as well. LCM(a, b) = (a x b) / GCD(a,b) or GCD(a,b) = (a x b) / LCM(a, b) 回答1: The analogous formulas to LCM(a, b) = (a x b) / GCD(a,b) or GCD(a,b) = (a x b) / LCM(a, b) with three variables are simply not valid, as your example with (3, 12, 10)

Least Common Multiple of an array values using Euclidean Algorithm

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问题 I want to calculate the least common multiple of an array of values, using Euclideans algorithm I am using this pseudocode implementation: found on wikipedia function gcd(a, b) while b ≠ 0 t := b; b := a mod b; a := t; return a; My javascript implementation is such function smallestCommons(arr) { var gcm = arr.reduce(function(a,b){ let minNum = Math.min(a,b); let maxNum = Math.max(a,b); var placeHolder = 0; while(minNum!==0){ placeHolder = maxNum; maxNum = minNum; minNum = placeHolder%minNum;

GCD and LCM relation

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The following relation works only for two (3, 12) numbers, it fails to produce the right answer when used for three numbers (3,12,10) . Just wondering if is it my understanding or it is just for two numbers and for me same is true for Euclid algorithm as well. LCM(a, b) = (a x b) / GCD(a,b) or GCD(a,b) = (a x b) / LCM(a, b) The analogous formulas to LCM(a, b) = (a x b) / GCD(a,b) or GCD(a,b) = (a x b) / LCM(a, b) with three variables are simply not valid, as your example with (3, 12, 10) shows readily. The product of these three numbers is 360. The GCD is 1. The LCM is 60. It is our common

LCM of all the numbers in an array in Java

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问题 I have an array of ints, and I'm trying to find the LCM (least common multiple) of all the values in the array. I've written an lcm method separately; it takes two values as input, and returns the lcm. My lcm method works perfectly fine, but when I use it to find the LCM of all the values I get a wrong answer. Here are my gcd and lcm methods: public static int gcd(int a, int b){ if (a<b) return gcd(b,a); if (a%b==0) return b; else return gcd(a, a%b); } public static int lcm(int a, int b){

LCM of all the numbers in an array in Java

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I have an array of ints, and I'm trying to find the LCM (least common multiple) of all the values in the array. I've written an lcm method separately; it takes two values as input, and returns the lcm. My lcm method works perfectly fine, but when I use it to find the LCM of all the values I get a wrong answer. Here are my gcd and lcm methods: public static int gcd(int a, int b){ if (a<b) return gcd(b,a); if (a%b==0) return b; else return gcd(a, a%b); } public static int lcm(int a, int b){ return ((a*b)/gcd(a,b)); } This is what I have for the lcm of the array values: public static int

C++ algorithm to calculate least common multiple for multiple numbers

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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) ? 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'; } boost provides functions for calculation lcm of 2 numbers (see here ) Then

LCM of n numbers modulo 1000000007

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问题 I have to find LCM of n numbers MODULO 10^9+7.My approach is find LCM of two numbers and then MOD them.Then take the LCM of next element and the answer obtained from the previous iteration and MOD them.Do this for all elements.Is it wrong ? 回答1: Yes, it is wrong. I have been working on a similar problem. You must be familiar with the following property of MOD: PROPERTY 1: (a*b*c*d...*p*q) % MOD = (a%MOD) (b%MOD) (c%MOD) ..... (q%MOD); but this is not the case with LCM. LCM(a,b)=a*b/GCD(a,b).

Finding the LCM of a range of numbers

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I read an interesting DailyWTF post today, "Out of All The Possible Answers..." and it interested me enough to dig up the original forum post where it was submitted. This got me thinking how I would solve this interesting problem - the original question is posed on Project Euler as: 2520 is the smallest number that can be divided by each of the numbers from 1 to 10 without any remainder. What is the smallest number that is evenly divisible by all of the numbers from 1 to 20? To reform this as a programming question, how would you create a function that can find the Least Common Multiple for an

C++ algorithm to calculate least common multiple for multiple numbers

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问题 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) ? 回答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);