分数

分数是两个值的比率。分数的形式为a / b，其中a称为分子，b称为分母，b不能等于0(因为未定义除以0)。分母给出了相等的部分。分子代表其中有多少个。例如，一半，八分之五，四分之三(1 / 2、8 / 5、3 / 4)。

关于分数的事实：

1. 如果分子和分母的最大公约数(gcd)大于1，则可以减小分数。
2. 分数的加法和减法：当分数相加或相减时，它们必须具有相同的分母。如果它们没有相同的分母，则我们必须为两者找到一个共同的分母。为此，我们首先需要找到两个分母的最低公倍数(lcm)或将每个分数乘以适当的整数，以便有相同的分母。
3. 分数的乘法和除法：将两个分数相乘时，只需将两个分子乘以两个分母即可。当将两个分数相除时，第一个分数必须与第二个分数的倒数相乘。
4. 分数分为三种：
   • 正确的分数：分子小于分母。例如1 / 3、3 / 4、2 / 7
   • 分数不正确：分子大于(或等于)分母。例如4 / 3、11 / 4、7 / 7。
   • 混合分数：整数和适当的分数在一起。例如1 1 / 3、2 1 / 4、16 2/5。

如何添加两个分数?
将两个分数a / b和c / d相加，并以最简单的形式打印答案。

例子 ：

Input:  1/2 + 3/2
Output: 2/1

Input:  1/3 + 3/9
Output: 2/3

Input:  1/5 + 3/15
Output: 2/5

加两个分数的算法

• 通过找到两个分母的LCM(最小公倍数)来找到一个公分母。
• 更改分数以具有相同的分母并添加两个术语。
• 通过将分子和分母除以最大的公因子，将获得的最终分数简化为更简单的形式。

C++

// C++ program to add 2 fractions
#include 
using namespace std;
  
// Function to return gcd of a and b
int gcd(int a, int b)
{
    if (a == 0)
        return b;
    return gcd(b % a, a);
}
  
// Function to convert the obtained fraction
// into it's simplest form
void lowest(int& den3, int& num3)
{
    // Finding gcd of both terms
    int common_factor = gcd(num3, den3);
  
    // Converting both terms into simpler
    // terms by dividing them by common factor
    den3 = den3 / common_factor;
    num3 = num3 / common_factor;
}
  
// Function to add two fractions
void addFraction(int num1, int den1, int num2,
                 int den2, int& num3, int& den3)
{
    // Finding gcd of den1 and den2
    den3 = gcd(den1, den2);
  
    // Denominator of final fraction obtained
    // finding LCM of den1 and den2
    // LCM * GCD = a * b
    den3 = (den1 * den2) / den3;
  
    // Changing the fractions to have same denominator
    // Numerator of the final fraction obtained
    num3 = (num1) * (den3 / den1) + (num2) * (den3 / den2);
  
    // Calling function to convert final fraction
    // into it's simplest form
    lowest(den3, num3);
}
  
// Driver program
int main()
{
    int num1 = 1, den1 = 500, num2 = 2, den2 = 1500, den3, num3;
  
    addFraction(num1, den1, num2, den2, num3, den3);
  
    printf("%d/%d + %d/%d is equal to %d/%d\n", num1, den1,
           num2, den2, num3, den3);
    return 0;
}

Java

// Java program to add 2 fractions
import java.util.*;
  
class GFG
{
static int den3, num3; 
  
// Function to return gcd of a and b
static int gcd(int a, int b)
{
    if (a == 0)
        return b;
    return gcd(b % a, a);
}
  
// Function to convert the obtained fraction
// into it's simplest form
static void lowest()
{
    // Finding gcd of both terms
    int common_factor = gcd(num3, den3);
  
    // Converting both terms into simpler
    // terms by dividing them by common factor
    den3 = den3 / common_factor;
    num3 = num3 / common_factor;
}
  
// Function to add two fractions
static void addFraction(int num1, int den1,
                        int num2, int den2)
{
    // Finding gcd of den1 and den2
    den3 = gcd(den1, den2);
  
    // Denominator of final fraction obtained
    // finding LCM of den1 and den2
    // LCM * GCD = a * b
    den3 = (den1 * den2) / den3;
  
    // Changing the fractions to have 
    // same denominator. 
    // Numerator of the final fraction obtained
    num3 = (num1) * (den3 / den1) + 
           (num2) * (den3 / den2);
  
    // Calling function to convert final fraction
    // into it's simplest form
    lowest();
}
  
// Driver Code
public static void main(String[] args) 
{
    int num1 = 1, den1 = 500,
        num2 = 2, den2 = 1500;
  
    addFraction(num1, den1, num2, den2);
  
    System.out.printf("%d/%d + %d/%d is equal to %d/%d\n", 
                      num1, den1, num2, den2, num3, den3);
}
}
  
// This code is contributed by Rajput-Ji

Python3

# Python3 program to add 2 fractions 
  
# Function to return gcd of a and b 
def gcd(a, b):
    if (a == 0): 
        return b 
    return gcd(b % a, a) 
  
# Function to convert the obtained 
# fraction into it's simplest form 
def lowest(den3, num3): 
  
    # Finding gcd of both terms 
    common_factor = gcd(num3, den3) 
  
    # Converting both terms 
    # into simpler terms by 
    # dividing them by common factor 
    den3 = int(den3 / common_factor) 
    num3 = int(num3 / common_factor)
    print(num3, "/", den3)
  
# Function to add two fractions 
def addFraction(num1, den1, num2, den2): 
  
    # Finding gcd of den1 and den2 
    den3 = gcd(den1, den2) 
  
    # Denominator of final 
    # fraction obtained finding 
    # LCM of den1 and den2 
    # LCM * GCD = a * b 
    den3 = (den1 * den2) / den3 
  
    # Changing the fractions to 
    # have same denominator Numerator 
    # of the final fraction obtained 
    num3 = ((num1) * (den3 / den1) +
            (num2) * (den3 / den2)) 
  
    # Calling function to convert 
    # final fraction into it's 
    # simplest form 
    lowest(den3, num3)
  
# Driver Code 
num1 = 1; den1 = 500
num2 = 2; den2 = 1500
  
print(num1, "/", den1, " + ", num2, "/", 
    den2, " is equal to ", end = "")
      
addFraction(num1, den1, num2, den2)

C#

// C# program to add 2 fractions
using System;
      
class GFG
{
static int den3, num3; 
  
// Function to return gcd of a and b
static int gcd(int a, int b)
{
    if (a == 0)
        return b;
    return gcd(b % a, a);
}
  
// Function to convert the obtained fraction
// into it's simplest form
static void lowest()
{
    // Finding gcd of both terms
    int common_factor = gcd(num3, den3);
  
    // Converting both terms into simpler
    // terms by dividing them by common factor
    den3 = den3 / common_factor;
    num3 = num3 / common_factor;
}
  
// Function to add two fractions
static void addFraction(int num1, int den1,
                        int num2, int den2)
{
    // Finding gcd of den1 and den2
    den3 = gcd(den1, den2);
  
    // Denominator of final fraction obtained
    // finding LCM of den1 and den2
    // LCM * GCD = a * b
    den3 = (den1 * den2) / den3;
  
    // Changing the fractions to have 
    // same denominator. 
    // Numerator of the final fraction obtained
    num3 = (num1) * (den3 / den1) + 
           (num2) * (den3 / den2);
  
    // Calling function to convert final fraction
    // into it's simplest form
    lowest();
}
  
// Driver Code
public static void Main(String[] args) 
{
    int num1 = 1, den1 = 500,
        num2 = 2, den2 = 1500;
  
    addFraction(num1, den1, num2, den2);
  
    Console.Write("{0}/{1} + {2}/{3} is equal to {4}/{5}\n", 
                        num1, den1, num2, den2, num3, den3);
}
}
  
// This code is contributed by PrinciRaj1992

输出 ：

1/500 + 2/1500 is equal to 1/300

与分数有关的更多问题：

• 馏分的LCM和HCF
• 以字符串格式表示两个数字的分数
• 程序比较两个分数
• 将二进制分数转换为十进制
• 将小数转换为二进制数
• 小背包问题
• 查找分数中的重复序列

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