求第一个数组的倍数和第二个数组的因子的数
原文:https://www . geesforgeks . org/find-numbers-它是第一个数组和第二个数组的倍数/
给定两个数组 A[] 和 B[] ，任务是找出能被数组 A[] 的所有元素整除的整数，再将数组 B[] 的所有元素整除。
示例:
输入: A[] = {1，2，2，4}，B[] = {16，32，64} 输出: 4 8 16 4，8，16 是唯一的数字 是数组 A[] 所有元素的倍数，除了数组 B[]
输入: A[] = {2，3，6}，B[] = {42，84 } T3】输出: 6 42
方法:如果 X 是第一个数组所有元素的倍数，那么 X 必须是第一个数组所有元素的 LCM 的倍数。 同样，如果 X 是第二个数组所有元素的因子，那么它一定是第二个数组所有元素的 GCD 的因子，这样的 X 只有在第二个数组的 GCD 被第一个数组的 LCM 整除的情况下才会存在。 如果可以整除，那么 X 可以是范围【LCM，GCD】中的任意值，该范围是 LCM 的倍数，并且均匀地划分 GCD。
下面是上述方法的实现:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// Function to return the LCM of two numbers
int lcm(int x, int y)
{
int temp = (x * y) / __gcd(x, y);
return temp;
}
// Function to print the required numbers
void findNumbers(int a[], int n, int b[], int m)
{
// To store the lcm of array a[] elements
// and the gcd of array b[] elements
int lcmA = 1, gcdB = 0;
// Finding LCM of first array
for (int i = 0; i < n; i++)
lcmA = lcm(lcmA, a[i]);
// Finding GCD of second array
for (int i = 0; i < m; i++)
gcdB = __gcd(gcdB, b[i]);
// No such element exists
if (gcdB % lcmA != 0) {
cout << "-1";
return;
}
// All the multiples of lcmA which are
// less than or equal to gcdB and evenly
// divide gcdB will satisfy the conditions
int num = lcmA;
while (num <= gcdB) {
if (gcdB % num == 0)
cout << num << " ";
num += lcmA;
}
}
// Driver code
int main()
{
int a[] = { 1, 2, 2, 4 };
int b[] = { 16, 32, 64 };
int n = sizeof(a) / sizeof(a[0]);
int m = sizeof(b) / sizeof(b[0]);
findNumbers(a, n, b, m);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java implementation of the approach
import java.util.*;
class GFG
{
static int __gcd(int a, int b)
{
if (b == 0)
return a;
return __gcd(b, a % b);
}
// Function to return the LCM of two numbers
static int lcm(int x, int y)
{
int temp = (x * y) / __gcd(x, y);
return temp;
}
// Function to print the required numbers
static void findNumbers(int a[], int n,
int b[], int m)
{
// To store the lcm of array a[] elements
// and the gcd of array b[] elements
int lcmA = 1, gcdB = 0;
// Finding LCM of first array
for (int i = 0; i < n; i++)
lcmA = lcm(lcmA, a[i]);
// Finding GCD of second array
for (int i = 0; i < m; i++)
gcdB = __gcd(gcdB, b[i]);
// No such element exists
if (gcdB % lcmA != 0)
{
System.out.print("-1");
return;
}
// All the multiples of lcmA which are
// less than or equal to gcdB and evenly
// divide gcdB will satisfy the conditions
int num = lcmA;
while (num <= gcdB)
{
if (gcdB % num == 0)
System.out.print(num + " ");
num += lcmA;
}
}
// Driver code
public static void main(String[] args)
{
int a[] = { 1, 2, 2, 4 };
int b[] = { 16, 32, 64 };
int n = a.length;
int m = b.length;
findNumbers(a, n, b, m);
}
}
// This code is contributed by 29AjayKumar
Python 3
# Python3 implementation of the approach
from math import gcd
# Function to return the LCM of two numbers
def lcm( x, y) :
temp = (x * y) // gcd(x, y);
return temp;
# Function to print the required numbers
def findNumbers(a, n, b, m) :
# To store the lcm of array a[] elements
# and the gcd of array b[] elements
lcmA = 1; __gcdB = 0;
# Finding LCM of first array
for i in range(n) :
lcmA = lcm(lcmA, a[i]);
# Finding GCD of second array
for i in range(m) :
__gcdB = gcd(__gcdB, b[i]);
# No such element exists
if (__gcdB % lcmA != 0) :
print("-1");
return;
# All the multiples of lcmA which are
# less than or equal to gcdB and evenly
# divide gcdB will satisfy the conditions
num = lcmA;
while (num <= __gcdB) :
if (__gcdB % num == 0) :
print(num, end = " ");
num += lcmA;
# Driver code
if __name__ == "__main__" :
a = [ 1, 2, 2, 4 ];
b = [ 16, 32, 64 ];
n = len(a);
m = len(b);
findNumbers(a, n, b, m);
# This code is contributed by AnkitRai01
C
// C# implementation of the approach
using System;
class GFG
{
static int __gcd(int a, int b)
{
if (b == 0)
return a;
return __gcd(b, a % b);
}
// Function to return the LCM of two numbers
static int lcm(int x, int y)
{
int temp = (x * y) / __gcd(x, y);
return temp;
}
// Function to print the required numbers
static void findNumbers(int []a, int n,
int []b, int m)
{
// To store the lcm of array a[] elements
// and the gcd of array b[] elements
int lcmA = 1, gcdB = 0;
// Finding LCM of first array
for (int i = 0; i < n; i++)
lcmA = lcm(lcmA, a[i]);
// Finding GCD of second array
for (int i = 0; i < m; i++)
gcdB = __gcd(gcdB, b[i]);
// No such element exists
if (gcdB % lcmA != 0)
{
Console.Write("-1");
return;
}
// All the multiples of lcmA which are
// less than or equal to gcdB and evenly
// divide gcdB will satisfy the conditions
int num = lcmA;
while (num <= gcdB)
{
if (gcdB % num == 0)
Console.Write(num + " ");
num += lcmA;
}
}
// Driver code
public static void Main(String[] args)
{
int []a = { 1, 2, 2, 4 };
int []b = { 16, 32, 64 };
int n = a.Length;
int m = b.Length;
findNumbers(a, n, b, m);
}
}
// This code is contributed by 29AjayKumar
java 描述语言
<script>
// Javascript implementation of the approach
// Function to find nth centered
// tridecagonal number
function __gcd(a, b)
{
if (b == 0)
return a;
return __gcd(b, a % b);
}
// Function to return the LCM of two numbers
function lcm(x, y)
{
var temp = (x * y) / __gcd(x, y);
return temp;
}
// Function to print the required numbers
function findNumbers(a, n, b, m)
{
// To store the lcm of array a[] elements
// and the gcd of array b[] elements
var lcmA = 1, gcdB = 0;
// Finding LCM of first array
for(var i = 0; i < n; i++)
lcmA = lcm(lcmA, a[i]);
// Finding GCD of second array
for(var i = 0; i < m; i++)
gcdB = __gcd(gcdB, b[i]);
// No such element exists
if (gcdB % lcmA != 0)
{
document.write("-1");
return;
}
// All the multiples of lcmA which are
// less than or equal to gcdB and evenly
// divide gcdB will satisfy the conditions
var num = lcmA;
while (num <= gcdB)
{
if (gcdB % num == 0)
document.write(num + " ");
num += lcmA;
}
}
// Driver code
var a = [ 1, 2, 2, 4 ];
var b = [ 16, 32, 64 ];
var n = a.length;
var m = b.length;
findNumbers(a, n, b, m);
// This code is contributed by Ankita saini
</script>
Output:
4 8 16
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