检查一个数组中完美平方的和是否能被 x 整除
原文:https://www . geeksforgeeks . org/check-如果数组中的完美平方和被 x 整除/
给定一个数组 arr[] 和一个整数 x ，任务是检查数组中所有完美平方的和是否能被 x 整除。如果可分，则打印是否则打印否。 举例:
输入: arr[] = {2，3，4，6，9，10}，x = 13 输出:是 4 和 9 是数组中唯一的完美正方形 和= 4 + 9 = 13(可被 13 整除) 输入:arr[= { 2，4，25，49，3，8}，x = 9 输出:否
方法:从 i 到n–1运行一个循环，检查arr【I】是否是完美的正方形。如果 arr[i] 是一个完美的正方形，那么更新 sum = sum + arr[i] 。如果最后总和% x = 0 则打印是否则打印否。要检查一个元素是否是完美的正方形，请遵循以下步骤:
假设 num 是一个整数元素 float sq = sqrt(x) 如果 floor(sq) = ceil(sq) 那么 num 就是一个完美的正方形，否则就不是。
以下是上述方法的实现:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// Function that returns true if the sum of all the
// perfect squares of the given array are divisible by x
bool check(int arr[], int x, int n)
{
long long sum = 0;
for (int i = 0; i < n; i++) {
double x = sqrt(arr[i]);
// If arr[i] is a perfect square
if (floor(x) == ceil(x)) {
sum += arr[i];
}
}
if (sum % x == 0)
return true;
else
return false;
}
// Driver code
int main()
{
int arr[] = { 2, 3, 4, 9, 10 };
int n = sizeof(arr) / sizeof(arr[0]);
int x = 13;
if (check(arr, x, n)) {
cout << "Yes";
}
else {
cout << "No";
}
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java implementation of the approach
public class GFG{
// Function that returns true if the the sum of all the
// perfect squares of the given array is divisible by x
static boolean check(int arr[], int x, int n)
{
long sum = 0;
for (int i = 0; i < n; i++) {
double y = Math.sqrt(arr[i]);
// If arr[i] is a perfect square
if (Math.floor(y) == Math.ceil(y)) {
sum += arr[i];
}
}
if (sum % x == 0)
return true;
else
return false;
}
// Driver Code
public static void main(String []args){
int arr[] = { 2, 3, 4, 9, 10 };
int n = arr.length ;
int x = 13;
if (check(arr, x, n)) {
System.out.println("Yes");
}
else {
System.out.println("No");
}
}
// This code is contributed by Ryuga
}
Python 3
# Python3 implementation of the approach
import math
# Function that returns true if the the sum of all the
# perfect squares of the given array is divisible by x
def check (a, y):
sum = 0
for i in range(len(a)):
x = math.sqrt(a[i])
# If a[i] is a perfect square
if (math.floor(x) == math.ceil(x)):
sum = sum + a[i]
if (sum % y == 0):
return True
else:
return False
# Driver code
a = [2, 3, 4, 9, 10]
x = 13
if check(a, x) :
print("Yes")
else:
print("No")
C
// C# implementation of the approach
using System;
public class GFG{
// Function that returns true if the the sum of all the
// perfect squares of the given array is divisible by x
static bool check(int[] arr, int x, int n)
{
long sum = 0;
for (int i = 0; i < n; i++) {
double y = Math.Sqrt(arr[i]);
// If arr[i] is a perfect square
if (Math.Floor(y) == Math.Ceiling(y)) {
sum += arr[i];
}
}
if (sum % x == 0)
return true;
else
return false;
}
// Driver Code
public static void Main(){
int[] arr = { 2, 3, 4, 9, 10 };
int n = arr.Length ;
int x = 13;
if (check(arr, x, n)) {
Console.Write("Yes");
}
else {
Console.Write("No");
}
}
}
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP implementation of the approach
// Function that returns true if the
// sum of all the perfect squares of
// the given array is divisible by x
function check($arr, $x, $n)
{
$sum = 0;
for ($i = 0; $i < $n; $i++)
{
$x = sqrt($arr[$i]);
// If arr[i] is a perfect square
if (floor($x) == ceil($x))
{
$sum += $arr[$i];
}
}
if (($sum % $x) == 0)
return true;
else
return false;
}
// Driver code
$arr = array( 2, 3, 4, 9, 10 );
$n = sizeof($arr);
$x = 13;
if (!check($arr, $x, $n))
{
echo "Yes";
}
else
{
echo "No";
}
// This code is contributed by Sachin
?>
java 描述语言
<script>
// Javascript implementation of the approach
// Function that returns true if the the sum of all the
// perfect squares of the given array is divisible by x
function check(arr,x,n)
{
let sum = 0;
for (let i = 0; i < n; i++) {
let y = Math.sqrt(arr[i]);
// If arr[i] is a perfect square
if (Math.floor(y) == Math.ceil(y)) {
sum += arr[i];
}
}
if (sum % x == 0)
return true;
else
return false;
}
// Driver Code
let arr=[ 2, 3, 4, 9, 10];
let n = arr.length ;
let x = 13;
if (check(arr, x, n)) {
document.write("Yes");
}
else {
document.write("No");
}
// This code is contributed by unknown2108
</script>
Output:
Yes
版权属于：月萌API www.moonapi.com，转载请注明出处
