馏分的 LCM 和 HCF
给定 n 个分数作为两个数组 Num 和 Den 。任务是找出分数的 L.C.M 。
示例:
输入: num[] = {1，7，4}，den[] = {2，3，6} 输出: LCM 为= 28/1 给定分数为 1/2，7/3 和 4/6。 LCM 为 28/1
输入: num[] = {24，48，72，96}，den[] = {2，6，8，3 } T3】输出: LCM 为= 288/1
A/B 和 C/D 的 LCM =(A 和 C 的 LCM)/(B 和 D 的 HCF)
下面是上述方法的实现:
C++
// C++ program to find LCM of array of fractions
#include <bits/stdc++.h>
using namespace std;
// Function that will calculate
// the Lcm of Numerator
int LCM(int num[], int N)
{
int ans = num[0];
for (int i = 1; i < N; i++)
ans = (((num[i] * ans)) / (__gcd(num[i], ans)));
return ans;
}
// Function that will calculate
// the Hcf of Denominator
int HCF(int den[], int N)
{
int ans = den[0];
for(int i = 1; i < N; i++)
ans = __gcd(den[i], ans);
return ans;
}
int LCMOfFractions(int num[], int den[], int N)
{
int Numerator = LCM(num, N);
int Denominator = HCF(den, N);
int gcd = __gcd(Numerator, Denominator);
Numerator = Numerator / gcd;
Denominator = Denominator / gcd;
cout << "LCM is = " << Numerator << "/" << Denominator;
}
// Driver code
int main()
{
int num[] = { 1, 7, 4 }, den[] = { 2, 3, 6 };
int N = sizeof(num) / sizeof(num[0]);
LCMOfFractions(num, den, N);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program to find LCM of array of fractions
class GFG{
// Recursive function to return gcd of a and b
static int gcd(int a, int b)
{
// Everything divides 0
if (a == 0)
return b;
if (b == 0)
return a;
// base case
if (a == b)
return a;
// a is greater
if (a > b)
return gcd(a-b, b);
return gcd(a, b-a);
}
// Function that will calculate
// the Lcm of Numerator
static int LCM(int num[], int N)
{
int ans = num[0];
for (int i = 1; i < N; i++)
ans = (((num[i] * ans)) / (gcd(num[i], ans)));
return ans;
}
// Function that will calculate
// the Hcf of Denominator
static int HCF(int den[], int N)
{
int ans = den[0];
for(int i = 1; i < N; i++)
ans = gcd(den[i], ans);
return ans;
}
static int LCMOfFractions(int num[], int den[], int N)
{
int Numerator = LCM(num, N);
int Denominator = HCF(den, N);
int gcd1 = gcd(Numerator, Denominator);
Numerator = Numerator / gcd1;
Denominator = Denominator / gcd1;
System.out.println("LCM is = " +Numerator+ "/" + Denominator);
return 0;
}
// Driver code
public static void main(String args[])
{
int num[] = { 1, 7, 4 }, den[] = { 2, 3, 6 };
int N = num.length;
LCMOfFractions(num, den, N);
}
}
Python 3
# Python3 def program to find LCM of
# array of fractions
# Recursive function to
# return gcd of a and b
def gcd(a, b):
# Everything divides 0
if (a == 0):
return b;
if (b == 0):
return a;
# base case
if (a == b):
return a;
# a is greater
if (a > b):
return gcd(a - b, b);
return gcd(a, b - a);
# Function that will calculate
# the Lcm of Numerator
def LCM(num, N):
ans = num[0];
for i in range(1,N):
ans = (((num[i] * ans)) / (gcd(num[i], ans)));
return ans;
# Function that will calculate
# the Hcf of Denominator
def HCF(den, N):
ans = den[0];
for i in range(1,N):
ans = gcd(den[i], ans);
return ans;
def LCMOfFractions(num, den, N):
Numerator = LCM(num, N);
Denominator = HCF(den, N);
gcd1 = gcd(Numerator, Denominator);
Numerator = int(Numerator / gcd1);
Denominator = int(Denominator / gcd1);
print("LCM is =",Numerator,"/",Denominator);
# Driver code
num = [1, 7, 4 ];
den = [2, 3, 6 ];
N = len(num);
LCMOfFractions(num, den, N);
# This code is contributed
# by mits
C
// C# program to find LCM of
// array of fractions
using System;
class GFG
{
// Recursive function to return
// gcd of a and b
static int gcd(int a, int b)
{
// Everything divides 0
if (a == 0)
return b;
if (b == 0)
return a;
// base case
if (a == b)
return a;
// a is greater
if (a > b)
return gcd(a - b, b);
return gcd(a, b - a);
}
// Function that will calculate
// the Lcm of Numerator
static int LCM(int []num, int N)
{
int ans = num[0];
for (int i = 1; i < N; i++)
ans = (((num[i] * ans)) /
(gcd(num[i], ans)));
return ans;
}
// Function that will calculate
// the Hcf of Denominator
static int HCF(int []den, int N)
{
int ans = den[0];
for(int i = 1; i < N; i++)
ans = gcd(den[i], ans);
return ans;
}
static int LCMOfFractions(int []num,
int []den, int N)
{
int Numerator = LCM(num, N);
int Denominator = HCF(den, N);
int gcd1 = gcd(Numerator, Denominator);
Numerator = Numerator / gcd1;
Denominator = Denominator / gcd1;
Console.WriteLine("LCM is = " + Numerator +
"/" + Denominator);
return 0;
}
// Driver code
static public void Main(String []args)
{
int[] num = { 1, 7, 4 }, den = { 2, 3, 6 };
int N = num.Length;
LCMOfFractions(num, den, N);
}
}
// This code is contributed by Arnab Kundu
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP program to find LCM of
// array of fractions
// Recursive function to
// return gcd of a and b
function gcd($a, $b)
{
// Everything divides 0
if ($a == 0)
return $b;
if ($b == 0)
return $a;
// base case
if ($a == $b)
return $a;
// a is greater
if ($a > $b)
return gcd($a - $b, $b);
return gcd($a, $b - $a);
}
// Function that will calculate
// the Lcm of Numerator
function LCM($num, $N)
{
$ans = $num[0];
for ($i = 1; $i < $N; $i++)
$ans = ((($num[$i] * $ans)) /
(gcd($num[$i], $ans)));
return $ans;
}
// Function that will calculate
// the Hcf of Denominator
function HCF($den, $N)
{
$ans = $den[0];
for($i = 1; $i < $N; $i++)
$ans = gcd($den[$i], $ans);
return $ans;
}
function LCMOfFractions($num, $den, $N)
{
$Numerator = LCM($num, $N);
$Denominator = HCF($den, $N);
$gcd1 = gcd($Numerator, $Denominator);
$Numerator = $Numerator / $gcd1;
$Denominator = $Denominator / $gcd1;
echo "LCM is = " . $Numerator .
"/" . $Denominator;
return 0;
}
// Driver code
$num = array(1, 7, 4 );
$den = array(2, 3, 6 );
$N = sizeof($num);
LCMOfFractions($num, $den, $N);
// This code is contributed
// by Akanksha Rai
java 描述语言
<script>
// Javascript program to find LCM of
// array of fractions
var num = [ 1, 7, 4 ];
var den = [ 2, 3, 6 ];
// Recursive function to return
// gcd of a and b
function gcd(a, b)
{
// Everything divides 0
if (a == 0)
return b;
if (b == 0)
return a;
// Base case
if (a == b)
return a;
// a is greater
if (a > b)
return gcd(a - b, b);
return gcd(a, b - a);
}
// Function that will calculate
// the Lcm of Numerator
function LCM(num, N)
{
var ans = num[0];
for(var i = 1; i < N; i++)
ans = (((num[i] * ans)) /
(gcd(num[i], ans)));
return ans;
}
// Function that will calculate
// the Hcf of Denominator
function HCF(den, N)
{
var ans = den[0];
for(var i = 1; i < N; i++)
ans = gcd(den[i], ans);
return ans;
}
function LCMOfFractions(num, den, N)
{
var Numerator = LCM(num, N);
var Denominator = HCF(den, N);
var gcd1 = gcd(Numerator, Denominator);
Numerator = Numerator / gcd1;
Denominator = Denominator / gcd1;
document.write("LCM is = " + Numerator +
"/" + Denominator);
return 0;
}
// Driver code
var N = num.length;
LCMOfFractions(num, den, N);
// This code is contributed by Ankita saini
</script>
Output:
LCM is = 28/1
给定 n 个分数作为两个数组 Num 和 Den。任务是找出分数的升程。
输入: num[] = {1，7，4}，den[] = {2，3，6} 输出: HCF 为 1/6 给定分数为 1/2，7/3 和 4/6。 HCF 为 1/6
输入: num[] = {24，48，72，96}，den[] = {2，6，8，3 } T3】输出: HCF 为 1/1
甲乙丙丁的碳氢化合物=(甲乙丙丁的碳氢化合物)/(甲乙丙丁的碳氢化合物)
下面是上述方法的实现:
C++
// CPP program to find GCD of array of fractions
#include <bits/stdc++.h>
using namespace std;
// Function that will calculate
// the Lcm of Denominator
int LCM(int den[], int N)
{
int ans = den[0];
for (int i = 1; i < N; i++)
ans = (((den[i] * ans)) / (__gcd(den[i], ans)));
return ans;
}
// Function that will calculate
// the Hcf of Numerator
int HCF(int num[], int N)
{
int ans = num[0];
for (int i = 1; i < N; i++)
ans = __gcd(num[i], ans);
return ans;
}
int HCFOfFractions(int num[], int den[], int N)
{
int Numerator = HCF(num, N);
int Denominator = LCM(den, N);
int result = __gcd(Numerator, Denominator);
Numerator = Numerator / result;
Denominator = Denominator / result;
cout << "HCF is = " << Numerator << "/" << Denominator;
}
// Driver code
int main()
{
int num[] = { 24, 48, 72, 96 }, den[] = { 2, 6, 8, 3 };
int N = sizeof(num) / sizeof(num[0]);
HCFOfFractions(num, den, N);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program to find GCD of array of fractions
class GFG{
static int __gcd(int a, int b)
{
if (a == 0)
return b;
return __gcd(b % a, a);
}
// Function that will calculate
// the Lcm of Denominator
static int LCM(int den[], int N)
{
int ans = den[0];
for (int i = 1; i < N; i++)
ans = (((den[i] * ans)) / (__gcd(den[i], ans)));
return ans;
}
// Function that will calculate
// the Hcf of Numerator
static int HCF(int num[], int N)
{
int ans = num[0];
for (int i = 1; i < N; i++)
ans = __gcd(num[i], ans);
return ans;
}
static void HCFOfFractions(int num[], int den[], int N)
{
int Numerator = HCF(num, N);
int Denominator = LCM(den, N);
int result = __gcd(Numerator, Denominator);
Numerator = Numerator / result;
Denominator = Denominator / result;
System.out.println("HCF is = "+Numerator+"/"+Denominator);
}
// Driver code
public static void main(String[] args)
{
int num[] = { 24, 48, 72, 96 }, den[] = { 2, 6, 8, 3 };
int N = num.length;
HCFOfFractions(num, den, N);
}
}
// This code is contributed by mits
Python 3
# Python3 def program to find LCM
# of array of fractions
# Recursive function to
# return gcd of a and b
def gcd(a, b):
# Everything divides 0
if (a == 0):
return b;
if (b == 0):
return a;
# base case
if (a == b):
return a;
# a is greater
if (a > b):
return gcd(a - b, b);
return gcd(a, b - a);
# Function that will calculate
# the Lcm of Numerator
def LCM(den, N):
ans = den[0];
for i in range(1,N):
ans = (((den[i] * ans)) /
(gcd(den[i], ans)));
return ans;
# Function that will calculate
# the Hcf of Denominator
def HCF(num, N):
ans = num[0];
for i in range(1, N):
ans = gcd(num[i], ans);
return ans;
def HCFOfFractions(num, den, N):
Numerator = HCF(num, N);
Denominator = LCM(den, N);
gcd1 = gcd(Numerator, Denominator);
Numerator = int(Numerator / gcd1);
Denominator = int(Denominator / gcd1);
print("HCF is =", Numerator,
"/", Denominator);
# Driver code
num = [24, 48, 72, 96 ];
den = [2, 6, 8, 3 ];
N = len(num);
HCFOfFractions(num, den, N);
# This code is contributed
# by Akanksha Rai
C
// C# program to find GCD of array of fractions
using System;
class GFG{
static int __gcd(int a, int b)
{
if (a == 0)
return b;
return __gcd(b % a, a);
}
// Function that will calculate
// the Lcm of Denominator
static int LCM(int[] den, int N)
{
int ans = den[0];
for (int i = 1; i < N; i++)
ans = (((den[i] * ans)) / (__gcd(den[i], ans)));
return ans;
}
// Function that will calculate
// the Hcf of Numerator
static int HCF(int[] num, int N)
{
int ans = num[0];
for (int i = 1; i < N; i++)
ans = __gcd(num[i], ans);
return ans;
}
static void HCFOfFractions(int[] num, int[] den, int N)
{
int Numerator = HCF(num, N);
int Denominator = LCM(den, N);
int result = __gcd(Numerator, Denominator);
Numerator = Numerator / result;
Denominator = Denominator / result;
Console.WriteLine("HCF is = "+Numerator+"/"+Denominator);
}
// Driver code
public static void Main()
{
int[] num = { 24, 48, 72, 96 }, den = { 2, 6, 8, 3 };
int N = num.Length;
HCFOfFractions(num, den, N);
}
}
// This code is contributed by mits
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP program to find GCD of
// array of fractions
function __gcd($a, $b)
{
if ($a == 0)
return $b;
return __gcd($b % $a, $a);
}
// Function that will calculate
// the Lcm of Denominator
function LCM($den, $N)
{
$ans = $den[0];
for ($i = 1; $i < $N; $i++)
$ans = ((($den[$i] * $ans)) /
(__gcd($den[$i], $ans)));
return $ans;
}
// Function that will calculate
// the Hcf of Numerator
function HCF($num, $N)
{
$ans = $num[0];
for ($i = 1; $i < $N; $i++)
$ans = __gcd($num[$i], $ans);
return $ans;
}
function HCFOfFractions($num, $den, $N)
{
$Numerator = HCF($num, $N);
$Denominator = LCM($den, $N);
$result = __gcd($Numerator, $Denominator);
$Numerator = $Numerator / $result;
$Denominator = $Denominator / $result;
echo "HCF is = " . $Numerator .
"/" . $Denominator;
}
// Driver code
$num = array( 24, 48, 72, 96 );
$den = array( 2, 6, 8, 3 );
$N = count($num);
HCFOfFractions($num, $den, $N);
// This code is contributed by mits
?>
java 描述语言
<script>
// Javascript program to find GCD of array of fractions
const __gcd = (a, b) => {
if(a == 0){
return b;
}
return __gcd(b % a, a);
}
// Function that will calculate
// the Lcm of Denominator
const LCM = (den, N) => {
let ans = den[0];
for (var i = 1; i < N; i++)
ans = (((den[i] * ans)) /
(__gcd(den[i], ans)));
return ans;
}
// Function that will calculate
// the Hcf of Numerator
const HCF = (num, N) => {
let ans = num[0];
for (var i = 1; i < N; i++)
ans = __gcd(num[i], ans);
return ans;
}
const HCFOfFractions = (num, den, N) => {
let Numerator = HCF(num, N);
let Denominator = LCM(den, N);
let result = __gcd(Numerator, Denominator);
Numerator = Numerator / result;
Denominator = Denominator / result;
document.write(`HCF is = ${Numerator} / ${Denominator}`);
}
// Driver code
let num = [24, 48, 72, 96 ];
let den = [2, 6, 8, 3 ];
let N = num.length;
HCFOfFractions(num, den, N);
// This code is contributed by _saurabh_jaiswal
</script>
Output:
HCF is = 1/1
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