最小公分母(液晶显示器)
原文:https://www.geeksforgeeks.org/least-common-denominator-lcd/
最小公分母或最小公分母是一组分数分母的最小公倍数。
公分母:两个或两个以上分数的分母相同时。 最小公分母是所有公分母中最小的。 为什么我们需要液晶？ 简化了加减法和比分数。 公分母可以简单地通过乘以分母来计算。在这种情况下，3 * 6 = 18
但这可能并不总是最小公分母，因为在这种情况下，液晶显示器= 6，而不是 18。液晶显示器实际上是分母的液晶显示器。 例:
LCD for fractions 5/12 and 7/15 is 60.
We can write both fractions as 25/60 and
28/60 so that they can be added and
subtracted easily.
LCD for fractions 1/3 and 4/7 is 21.
例题:给定两个分数，用最小公约数求它们的和。 例 :
Input : 1/6 + 7/15
Output : 19/30
Explanation : LCM of 6 and 15 is 30\.
So, 5/30 + 14/30 = 19/30
Input : 1/3 + 1/6
Output : 3/6
Explanation : LCM of 3 and 6 is 6\.
So, 2/6 + 1/6 = 3/6
注* 这些答案可以通过反常抵消进一步简化。
C++
// C++ Program to determine
// LCD of two fractions and
// Perform addition on fractions
#include <iostream>
using namespace std;
// function to calculate gcd
// or hcf of two numbers.
int gcd(int a, int b)
{
if (a == 0)
return b;
return gcd(b % a, a);
}
// function to calculate
// lcm of two numbers.
int lcm(int a, int b)
{
return (a * b) / gcd(a, b);
}
void printSum(int num1, int den1,
int num2, int den2)
{
// least common multiple
// of denominators LCD
// of 6 and 15 is 30.
int lcd = lcm(den1, den2);
// Computing the numerators for LCD:
// Writing 1/6 as 5/30 and 7/15 as
// 14/30
num1 *= (lcd / den1);
num2 *= (lcd / den2);
// Our sum is going to be res_num/lcd
int res_num = num1 + num2;
cout << res_num << "/" << lcd;
}
// Driver Code
int main()
{
// First fraction is 1/6
int num1 = 1, den1 = 6;
// Second fraction is 7/15
int num2 = 7, den2 = 15;
printSum(num1, den1, num2, den2);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java Program to determine LCD of two
// fractions and Perform addition on
// fractions
public class GFG {
// function to calculate gcd or
// hcf of two numbers.
static int gcd(int a, int b)
{
if (a == 0)
return b;
return gcd(b % a, a);
}
// function to calculate lcm of
// two numbers.
static int lcm(int a, int b)
{
return (a * b) / gcd(a, b);
}
static void printSum(int num1, int den1,
int num2, int den2)
{
// least common multiple of
// denominators LCD of 6 and 15
// is 30.
int lcd = lcm(den1, den2);
// Computing the numerators for LCD:
// Writing 1/6 as 5/30 and 7/15 as
// 14/30
num1 *= (lcd / den1);
num2 *= (lcd / den2);
// Our sum is going to be res_num/lcd
int res_num = num1 + num2;
System.out.print( res_num + "/" + lcd);
}
// Driver code
public static void main(String args[])
{
// First fraction is 1/6
int num1 = 1, den1 = 6;
// Second fraction is 7/15
int num2 = 7, den2 = 15;
printSum(num1, den1, num2, den2);
}
}
// This code is contributed by Sam007.
Python 3
# python Program to determine
# LCD of two fractions and
# Perform addition on fractions
# function to calculate gcd
# or hcf of two numbers.
def gcd(a, b):
if (a == 0):
return b
return gcd(b % a, a)
# function to calculate
# lcm of two numbers.
def lcm(a, b):
return (a * b) / gcd(a, b)
def printSum(num1, den1,
num2, den2):
# least common multiple
# of denominators LCD
# of 6 and 15 is 30.
lcd = lcm(den1, den2);
# Computing the numerators
# for LCD: Writing 1/6 as
# 5/30 and 7/15 as 14/30
num1 *= (lcd / den1)
num2 *= (lcd / den2)
# Our sum is going to be
# res_num/lcd
res_num = num1 + num2;
print( int(res_num) , "/" ,
int(lcd))
# Driver Code
# First fraction is 1/6
num1 = 1
den1 = 6
# Second fraction is 7/15
num2 = 7
den2 = 15
printSum(num1, den1, num2, den2);
# This code is contributed
# by Sam007
C
// C# Program to determine LCD of two
// fractions and Perform addition on
// fractions
using System;
class GFG {
// function to calculate gcd or
// hcf of two numbers.
static int gcd(int a, int b)
{
if (a == 0)
return b;
return gcd(b % a, a);
}
// function to calculate lcm of
// two numbers.
static int lcm(int a, int b)
{
return (a * b) / gcd(a, b);
}
static void printSum(int num1, int den1,
int num2, int den2)
{
// least common multiple of
// denominators LCD of 6 and 15
// is 30.
int lcd = lcm(den1, den2);
// Computing the numerators for LCD:
// Writing 1/6 as 5/30 and 7/15 as
// 14/30
num1 *= (lcd / den1);
num2 *= (lcd / den2);
// Our sum is going to be res_num/lcd
int res_num = num1 + num2;
Console.Write( res_num + "/" + lcd);
}
// Driver code
public static void Main ()
{
// First fraction is 1/6
int num1 = 1, den1 = 6;
// Second fraction is 7/15
int num2 = 7, den2 = 15;
printSum(num1, den1, num2, den2);
}
}
// This code is contributed by Sam007.
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP Program to determine
// LCD of two fractions and
// Perform addition on fractions
// function to calculate gcd
// or hcf of two numbers.
function gcd($a,$b)
{
if ($a == 0)
return $b;
return gcd($b % $a, $a);
}
// function to calculate
// lcm of two numbers.
function lcm($a,$b)
{
return ($a * $b) / gcd($a, $b);
}
function printSum($num1, $den1,
$num2, $den2)
{
// least common multiple
// of denominators
// LCD of 6 and 15 is 30.
$lcd = lcm($den1, $den2);
// Computing the numerators for LCD:
// Writing 1/6 as 5/30 and 7/15 as
// 14/30
$num1 *= ($lcd / $den1);
$num2 *= ($lcd / $den2);
// Our sum is going to be res_num/lcd
$res_num = $num1 + $num2;
echo $res_num . "/" . $lcd;
}
// Driver Code
// First fraction is 1/6
$num1 = 1;
$den1 = 6;
// Second fraction is 7/15
$num2 = 7;
$den2 = 15;
printSum($num1, $den1, $num2, $den2);
// This code is contributed by Sam007.
?>
java 描述语言
<script>
// javascript Program to determine LCD of two
// fractions and Perform addition on
// fractions
// function to calculate gcd or
// hcf of two numbers.
function gcd(a , b)
{
if (a == 0)
return b;
return gcd(b % a, a);
}
// function to calculate lcm of
// two numbers.
function lcm(a , b)
{
return (a * b) / gcd(a, b);
}
function printSum(num1 , den1 , num2 , den2)
{
// least common multiple of
// denominators LCD of 6 and 15
// is 30.
var lcd = lcm(den1, den2);
// Computing the numerators for LCD:
// Writing 1/6 as 5/30 and 7/15 as
// 14/30
num1 *= (lcd / den1);
num2 *= (lcd / den2);
// Our sum is going to be res_num/lcd
var res_num = num1 + num2;
document.write(res_num + "/" + lcd);
}
// Driver code
// First fraction is 1/6
var num1 = 1, den1 = 6;
// Second fraction is 7/15
var num2 = 7, den2 = 15;
printSum(num1, den1, num2, den2);
// This code is contributed by todaysgaurav
</script>
输出:
19/30
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