给定 N 个简化形式分数的乘积
原文:https://www . geesforgeks . org/product-given-n-fractions-reduced-form/
给定 N 个分数的分子和分母。任务是找到 N 分数的乘积，并以简化形式输出答案。 例:
Input : N = 3
num[] = { 1, 2, 5 }
den[] = { 2, 1, 6 }
Output : 5/6
Product of 1/2 * 2/1 * 5/6 is 10/12.
Reduced form of 10/12 is 5/6.
Input : N = 4
num[] = { 1, 2, 5, 9 }
den[] = { 2, 1, 6, 27 }
Output : 5/18
想法是在一个变量中找到分子的乘积，比如 new_num。现在，在另一个变量中找到分母的乘积，比如 new_den。 现在，要以约简形式求答案，求 new_num 和 new_den 的最大公约数，并用计算出的 GCD 除 new_num 和 new_den。 以下是本办法的实施:
C++
// CPP program to find product of N
// fractions in reduced form.
#include <bits/stdc++.h>
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);
}
// Print the Product of N fraction in Reduced Form.
void productReduce(int n, int num[], int den[])
{
int new_num = 1, new_den = 1;
// finding the product of all N
// numerators and denominators.
for (int i = 0; i < n; i++) {
new_num *= num[i];
new_den *= den[i];
}
// Finding GCD of new numerator and
// denominator
int GCD = gcd(new_num, new_den);
// Converting into reduced form.
new_num /= GCD;
new_den /= GCD;
cout << new_num << "/" << new_den << endl;
}
// Driven Program
int main()
{
int n = 3;
int num[] = { 1, 2, 5 };
int den[] = { 2, 1, 6 };
productReduce(n, num, den);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program to find product of N
// fractions in reduced form.
import java.io.*;
class GFG {
// 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);
}
// Print the Product of N fraction in
// Reduced Form.
static void productReduce(int n, int num[],
int den[])
{
int new_num = 1, new_den = 1;
// finding the product of all N
// numerators and denominators.
for (int i = 0; i < n; i++) {
new_num *= num[i];
new_den *= den[i];
}
// Finding GCD of new numerator and
// denominator
int GCD = gcd(new_num, new_den);
// Converting into reduced form.
new_num /= GCD;
new_den /= GCD;
System.out.println(new_num + "/" +new_den);
}
// Driven Program
public static void main (String[] args) {
int n = 3;
int num[] = { 1, 2, 5 };
int den[] = { 2, 1, 6 };
productReduce(n, num, den);
}
}
//This code is contributed by vt_m.
Python 3
# Python3 program to find
# product of N fractions
# in reduced form.
# Function to return
# gcd of a and b
def gcd(a, b):
if (a == 0):
return b;
return gcd(b % a, a);
# Print the Product of N
# fraction in Reduced Form.
def productReduce(n, num, den):
new_num = 1;
new_den = 1;
# finding the product
# of all N numerators
# and denominators.
for i in range(n):
new_num = new_num * num[i];
new_den = new_den * den[i];
# Finding GCD of
# new numerator
# and denominator
GCD = gcd(new_num, new_den);
# Converting into
# reduced form.
new_num = new_num / GCD;
new_den = new_den / GCD;
print(int(new_num), "/",
int(new_den));
# Driver Code
n = 3;
num = [1, 2, 5];
den = [2, 1, 6];
productReduce(n, num, den);
# This code is contributed
# by mits
C
// C# program to find product of N
// fractions in reduced form.
using System;
class GFG {
// 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);
}
// Print the Product of N fraction in
// Reduced Form.
static void productReduce(int n, int []num,
int []den)
{
int new_num = 1, new_den = 1;
// finding the product of all N
// numerators and denominators.
for (int i = 0; i < n; i++) {
new_num *= num[i];
new_den *= den[i];
}
// Finding GCD of new numerator and
// denominator
int GCD = gcd(new_num, new_den);
// Converting into reduced form.
new_num /= GCD;
new_den /= GCD;
Console.WriteLine(new_num + "/" +new_den);
}
// Driven Program
public static void Main () {
int n = 3;
int []num = { 1, 2, 5 };
int []den = { 2, 1, 6 };
productReduce(n, num, den);
}
}
//This code is contributed by vt_m.
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP program to find product of N
// fractions in reduced form.
// Function to return
// gcd of a and b
function gcd($a, $b)
{
if ($a == 0)
return $b;
return gcd($b % $a, $a);
}
// Print the Product of N
// fraction in Reduced Form.
function productReduce($n, $num, $den)
{
$new_num = 1; $new_den = 1;
// finding the product of all N
// numerators and denominators.
for ($i = 0; $i < $n; $i++)
{
$new_num *= $num[$i];
$new_den *= $den[$i];
}
// Finding GCD of new
// numerator and denominator
$GCD = gcd($new_num, $new_den);
// Converting into reduced form.
$new_num /= $GCD;
$new_den /= $GCD;
echo $new_num , "/" , $new_den ,"\n";
}
// Driver Code
$n = 3;
$num = array(1, 2, 5);
$den = array(2, 1, 6);
productReduce($n, $num, $den);
// This code is contributed by ajit
?>
java 描述语言
<script>
// JavaScript program to find product of N
// fractions in reduced form.
// Function to return gcd of a and b
function gcd( a, b)
{
if (a == 0)
return b;
return gcd(b % a, a);
}
// Print the Product of N fraction in Reduced Form.
function productReduce( n, num , den)
{
let new_num = 1, new_den = 1;
// finding the product of all N
// numerators and denominators.
for (let i = 0; i < n; i++) {
new_num *= num[i];
new_den *= den[i];
}
// Finding GCD of new numerator and
// denominator
let GCD = gcd(new_num, new_den);
// Converting into reduced form.
new_num /= GCD;
new_den /= GCD;
document.write( new_num + "/" + new_den);
}
// Driven Program
let n = 3;
let num = [ 1, 2, 5 ];
let den = [ 2, 1, 6 ];
productReduce(n, num, den);
// This code contributed by aashish1995
</script>
输出:
5/6
如何避免溢出？ 上述解决方案导致大量溢出。我们可以通过首先找到所有记数器和分母的质因数来避免溢出。一旦我们找到了素因子，我们就可以取消常见的素因子。 注意:当要求你以{ p \u q } ^ {-1 } }的形式表示答案时。 对于这些问题，首先将分子和分母以可约形式 P / Q 转换，如上所述。然后，求 q 相对于一个质数 m(一般是 10^9 + 7)的模乘逆。求 Q 的模乘逆后，与 P 相乘，取给定素数 m 的模，得到我们需要的输出。 //感谢 VaiBzZk 提示此情况。
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