分数(或有理数)数组的 HCF
给定一个分数序列。求给定分数系列的高频系数。 例:
Input : [{2, 5}, {8, 9}, {16, 81}, {10, 27}]
Output : 2, 405
Explanation : 2/405 is the largest number that
divides all 2/5, 8/9, 16/81 and 10/27.
Input : [{9, 10}, {12, 25}, {18, 35}, {21, 40}]
Output : 3, 1400
进场:
找到记数器的高频。 求分母的 L.C.M。 计算高频/低频交流电的分数 将分数降低至最低分数。
C++
// CPP program to find HCF of array of
// rational numbers (fractions).
#include <iostream>
using namespace std;
// hcf of two number
int gcd(int a, int b)
{
if (a % b == 0)
return b;
else
return (gcd(b, a % b));
}
// find hcf of numerator series
int findHcf(int** arr, int size)
{
int ans = arr[0][0];
for (int i = 1; i < size; i++)
ans = gcd(ans, arr[i][0]);
// return hcf of numerator
return (ans);
}
// find lcm of denominator series
int findLcm(int** arr, int size)
{
// ans contains LCM of arr[0][1], ..arr[i][1]
int ans = arr[0][1];
for (int i = 1; i < size; i++)
ans = (((arr[i][1] * ans)) /
(gcd(arr[i][1], ans)));
// return lcm of denominator
return (ans);
}
// Core Function
int* hcfOfFraction(int** arr, int size)
{
// found hcf of numerator
int hcf_of_num = findHcf(arr, size);
// found lcm of denominator
int lcm_of_deno = findLcm(arr, size);
int* result = new int[2];
result[0] = hcf_of_num;
result[1] = lcm_of_deno;
for (int i = result[0] / 2; i > 1; i--)
{
if ((result[1] % i == 0) && (result[0] % i == 0))
{
result[1] /= i;
result[0] /= i;
}
}
// return result
return (result);
}
// Main function
int main()
{
int size = 4;
int** arr = new int*[size];
// Initialize the every row
// with size 2 (1 for numerator
// and 2 for denominator)
for (int i = 0; i < size; i++)
arr[i] = new int[2];
arr[0][0] = 9;
arr[0][1] = 10;
arr[1][0] = 12;
arr[1][1] = 25;
arr[2][0] = 18;
arr[2][1] = 35;
arr[3][0] = 21;
arr[3][1] = 40;
// function for calculate the result
int* result = hcfOfFraction(arr, size);
// print the result
cout << result[0] << ", " << result[1] << endl;
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program to find HCF of array of
// rational numbers (fractions).
class GFG
{
// hcf of two number
static int gcd(int a, int b)
{
if (a % b == 0)
return b;
else
return (gcd(b, a % b));
}
// find hcf of numerator series
static int findHcf(int [][]arr, int size)
{
int ans = arr[0][0];
for (int i = 1; i < size; i++)
ans = gcd(ans, arr[i][0]);
// return hcf of numerator
return (ans);
}
// find lcm of denominator series
static int findLcm(int[][] arr, int size)
{
// ans contains LCM of arr[0][1], ..arr[i][1]
int ans = arr[0][1];
for (int i = 1; i < size; i++)
ans = (((arr[i][1] * ans)) /
(gcd(arr[i][1], ans)));
// return lcm of denominator
return (ans);
}
// Core Function
static int[] hcfOfFraction(int[][] arr, int size)
{
// found hcf of numerator
int hcf_of_num = findHcf(arr, size);
// found lcm of denominator
int lcm_of_deno = findLcm(arr, size);
int[] result = new int[2];
result[0] = hcf_of_num;
result[1] = lcm_of_deno;
for (int i = result[0] / 2; i > 1; i--)
{
if ((result[1] % i == 0) && (result[0] % i == 0))
{
result[1] /= i;
result[0] /= i;
}
}
// return result
return (result);
}
// Driver code
public static void main(String[] args)
{
int size = 4;
int[][] arr = new int[size][size];
// Initialize the every row
// with size 2 (1 for numerator
// and 2 for denominator)
for (int i = 0; i < size; i++)
arr[i] = new int[2];
arr[0][0] = 9;
arr[0][1] = 10;
arr[1][0] = 12;
arr[1][1] = 25;
arr[2][0] = 18;
arr[2][1] = 35;
arr[3][0] = 21;
arr[3][1] = 40;
// function for calculate the result
int[] result = hcfOfFraction(arr, size);
// print the result
System.out.println(result[0] + ", " + result[1]);
}
}
/* This code contributed by PrinciRaj1992 */
Python 3
# Python 3 program to find HCF of array of
from math import gcd
# find hcf of numerator series
def findHcf(arr, size):
ans = arr[0][0]
for i in range(1, size, 1):
ans = gcd(ans, arr[i][0])
# return hcf of numerator
return (ans)
# find lcm of denominator series
def findLcm(arr, size):
# ans contains LCM of arr[0][1], ..arr[i][1]
ans = arr[0][1]
for i in range(1, size, 1):
ans = int((((arr[i][1] * ans)) /
(gcd(arr[i][1], ans))))
# return lcm of denominator
return (ans)
# Core Function
def hcfOfFraction(arr, size):
# found hcf of numerator
hcf_of_num = findHcf(arr, size)
# found lcm of denominator
lcm_of_deno = findLcm(arr, size)
result = [0 for i in range(2)]
result[0] = hcf_of_num
result[1] = lcm_of_deno
i = int(result[0] / 2)
while(i > 1):
if ((result[1] % i == 0) and
(result[0] % i == 0)):
result[1] = int(result[1] / i)
result[0] = (result[0] / i)
# return result
return (result)
# Driver Code
if __name__ == '__main__':
size = 4
arr = [0 for i in range(size)]
# Initialize the every row
# with size 2 (1 for numerator
# and 2 for denominator)
for i in range(size):
arr[i] = [0 for i in range(2)]
arr[0][0] = 9
arr[0][1] = 10
arr[1][0] = 12
arr[1][1] = 25
arr[2][0] = 18
arr[2][1] = 35
arr[3][0] = 21
arr[3][1] = 40
# function for calculate the result
result = hcfOfFraction(arr, size)
# print the result
print(result[0], ",", result[1])
# This code is contributed by
# Surendra_Gangwar
C
// C# program to find HCF of array of
// rational numbers (fractions).
using System;
class GFG
{
// hcf of two number
static int gcd(int a, int b)
{
if (a % b == 0)
return b;
else
return (gcd(b, a % b));
}
// find hcf of numerator series
static int findHcf(int [,]arr, int size)
{
int ans = arr[0, 0];
for (int i = 1; i < size; i++)
ans = gcd(ans, arr[i, 0]);
// return hcf of numerator
return (ans);
}
// find lcm of denominator series
static int findLcm(int[,] arr, int size)
{
// ans contains LCM of arr[0,1], ..arr[i,1]
int ans = arr[0,1];
for (int i = 1; i < size; i++)
ans = (((arr[i, 1] * ans)) /
(gcd(arr[i, 1], ans)));
// return lcm of denominator
return (ans);
}
// Core Function
static int[] hcfOfFraction(int[,] arr, int size)
{
// found hcf of numerator
int hcf_of_num = findHcf(arr, size);
// found lcm of denominator
int lcm_of_deno = findLcm(arr, size);
int[] result = new int[2];
result[0] = hcf_of_num;
result[1] = lcm_of_deno;
for (int i = result[0] / 2; i > 1; i--)
{
if ((result[1] % i == 0) && (result[0] % i == 0))
{
result[1] /= i;
result[0] /= i;
}
}
// return result
return (result);
}
// Driver code
public static void Main(String[] args)
{
int size = 4;
int[,] arr = new int[size, size];
// Initialize the every row
// with size 2 (1 for numerator
// and 2 for denominator)
arr[0, 0] = 9;
arr[0, 1] = 10;
arr[1, 0] = 12;
arr[1, 1] = 25;
arr[2, 0] = 18;
arr[2, 1] = 35;
arr[3, 0] = 21;
arr[3, 1] = 40;
// function for calculate the result
int[] result = hcfOfFraction(arr, size);
// print the result
Console.WriteLine(result[0] + ", " + result[1]);
}
}
// This code has been contributed by 29AjayKumar
java 描述语言
<script>
// Javascript program to find HCF of array of
// rational numbers (fractions).
// hcf of two number
function gcd(a,b)
{
if (a % b == 0)
return b;
else
return (gcd(b, a % b));
}
// find hcf of numerator series
function findHcf(arr,size)
{
let ans = arr[0][0];
for (let i = 1; i < size; i++)
ans = gcd(ans, arr[i][0]);
// return hcf of numerator
return (ans);
}
// find lcm of denominator series
function findLcm(arr,size)
{
// ans contains LCM of arr[0][1], ..arr[i][1]
let ans = arr[0][1];
for (let i = 1; i < size; i++)
ans = Math.floor(((arr[i][1] * ans)) /
(gcd(arr[i][1], ans)));
// return lcm of denominator
return (ans);
}
// Core Function
function hcfOfFraction(arr,size)
{
// found hcf of numerator
let hcf_of_num = findHcf(arr, size);
// found lcm of denominator
let lcm_of_deno = findLcm(arr, size);
let result = new Array(2);
result[0] = hcf_of_num;
result[1] = lcm_of_deno;
for (let i = result[0] / 2; i > 1; i--)
{
if ((result[1] % i == 0) && (result[0] % i == 0))
{
result[1] /= i;
result[0] /= i;
}
}
// return result
return (result);
}
// Driver code
let size = 4;
let arr = new Array(size);
// Initialize the every row
// with size 2 (1 for numerator
// and 2 for denominator)
for (let i = 0; i < size; i++)
arr[i] = new Array(2);
arr[0][0] = 9;
arr[0][1] = 10;
arr[1][0] = 12;
arr[1][1] = 25;
arr[2][0] = 18;
arr[2][1] = 35;
arr[3][0] = 21;
arr[3][1] = 40;
// function for calculate the result
let result = hcfOfFraction(arr, size);
// print the result
document.write(result[0] + ", " + result[1]+"<br>");
// This code is contributed by rag2127
</script>
输出:
3, 1400
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