给定输入为分数，即 a/b 和 c/d，其中 a、b、c 和 d 可以是除 0 以外的任何整数值，任务是将这两个分数相加以生成它们的最终和。
分数用 − 表示
- a / b，其中 a 被称为分子，b 被称为分母。
- a 和 b 可以有任何数值，但 b 不能为 0。
- 两个分数的和表示为 a / b + c / d，添加这两个项的规则是它们的分母必须相等，如果它们不相等，则应使它们相等，然后才能进行相加。
示例
Input-: 1/4 + 2/12 Output-: 5/12 Since both the fractions denominators are unequal so to make them equal either GCD or LCM can be calculated. So in this case by multiplying the denominator which is 4 by 3 we can make them equal (1 * 3) / (4 * 3) = 3 / 12 Add both the terms: 3 / 12 + 2 / 12 = 5 / 12 Input-: 1/4 + 2/4 Output-: 3/4 Since both the terms have same denominator they can be directly added
算法
In function int gcd(int a, int b) Step 1-> If a == 0 then, return b Step 2-> Return gcd(b%a, a) In function void smallest(int &den3, int &n3) Step 1-> Declare and initialize common_factor as gcd(n3,den3) Step 2-> Set den3 = den3/common_factor Step 3-> Set n3 = n3/common_factor In Function void add_frac(int n1, int den1, int n2, int den2, int &n3, int &den3) Step 1-> Set den3 = gcd(den1,den2) Step 2-> Set den3 = (den1*den2) / den3 Step 3-> Set n3 = (n1)*(den3/den1) + (n2)*(den3/den2) Step 4-> Call function smallest(den3,n3) In Function int main() Step 1-> Declare and initialize n1=1, den1=4, n2=2, den2=12, den3, n3 Step 2-> Call add_frac(n1, den1, n2, den2, n3, den3) Step 3-> Print the values of n1, den1, n2, den2, n3, den3
Example
的中文翻译为：示例
#include <stdio.h>
int gcd(int a, int b) {
if (a == 0)
return b;
return gcd(b%a, a);
}
void smallest(int &den3, int &n3) {
// Finding gcd of both terms
int common_factor = gcd(n3,den3);
den3 = den3/common_factor;
n3 = n3/common_factor;
}
void add_frac(int n1, int den1, int n2, int den2, int &n3, int &den3) {
// to find the gcd of den1 and den2
den3 = gcd(den1,den2);
// LCM * GCD = a * b
den3 = (den1*den2) / den3;
// Changing the inputs to have same denominator
// Numerator of the final fraction obtained
n3 = (n1)*(den3/den1) + (n2)*(den3/den2);
smallest(den3,n3);
}
// Driver program
int main() {
int n1=1, den1=4, n2=2, den2=12, den3, n3;
add_frac(n1, den1, n2, den2, n3, den3);
printf("%d/%d + %d/%d = %d/%d</p><p>", n1, den1, n2, den2, n3, den3);
return 0;
}输出
1/4 + 2/12 = 5/12
