计算除所有其他元素之和的数组元素
原文:https://www . geesforgeks . org/count-array-elements-除以所有其他元素的和/
给定一个数组 arr[] ，任务是计算数组中除所有其他元素之和的元素数量。
示例:
输入: arr[] = {3，10，4，6，7} 输出: 3 3 除(10 + 4 + 6 + 7)即 27 10 除(3 + 4 + 6 + 7)即 20 6 除(3 + 10 + 4 + 7)即 24
输入: arr[] = {1，2，3，4，5，6，7，8，9，10} 输出: 2
天真方法:从 0 到 N 运行两个循环，计算除当前元素之外的所有元素的总和，如果该元素除以该总和，则增加计数。 以下是上述办法的实施情况:
C++
// C++ implementation of the approach
#include <iostream>
using namespace std;
// Function to return the count
// of the required numbers
int countNum(int N, int arr[])
{
// To store the count of required numbers
int count = 0;
for (int i = 0; i < N; i++) {
// Initialize sum to 0
int sum = 0;
for (int j = 0; j < N; j++) {
// If current element and the
// chosen element are same
if (i == j)
continue;
// Add all other numbers of array
else
sum += arr[j];
}
// If sum is divisible by the chosen element
if (sum % arr[i] == 0)
count++;
}
// Return the count
return count;
}
// Driver code
int main()
{
int arr[] = { 3, 10, 4, 6, 7 };
int n = sizeof(arr) / sizeof(arr[0]);
cout << countNum(n, arr);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java implementation of the approach
class GFG
{
// Function to return the count
// of the required numbers
static int countNum(int N, int arr[])
{
// To store the count of required numbers
int count = 0;
for (int i = 0; i < N; i++)
{
// Initialize sum to 0
int sum = 0;
for (int j = 0; j < N; j++)
{
// If current element and the
// chosen element are same
if (i == j)
continue;
// Add all other numbers of array
else
sum += arr[j];
}
// If sum is divisible by the chosen element
if (sum % arr[i] == 0)
count++;
}
// Return the count
return count;
}
// Driver code
public static void main(String[] args)
{
int arr[] = { 3, 10, 4, 6, 7 };
int n = arr.length;
System.out.println(countNum(n, arr));
}
}
// This code is contributed by Code_Mech
Python 3
# Python3 implementation of the approach
# Function to return the count
# of the required numbers
def countNum(N, arr):
# To store the count of
# required numbers
count = 0
for i in range(N):
# Initialize sum to 0
Sum = 0
for j in range(N):
# If current element and the
# chosen element are same
if (i == j):
continue
# Add all other numbers of array
else:
Sum += arr[j]
# If Sum is divisible by the
# chosen element
if (Sum % arr[i] == 0):
count += 1
# Return the count
return count
# Driver code
arr = [3, 10, 4, 6, 7]
n = len(arr)
print(countNum(n, arr))
# This code is contributed
# by Mohit Kumar
C
// C# implementation of the approach
using System;
class GFG
{
// Function to return the count
// of the required numbers
static int countNum(int N, int []arr)
{
// To store the count of required numbers
int count = 0;
for (int i = 0; i < N; i++)
{
// Initialize sum to 0
int sum = 0;
for (int j = 0; j < N; j++)
{
// If current element and the
// chosen element are same
if (i == j)
continue;
// Add all other numbers of array
else
sum += arr[j];
}
// If sum is divisible by the chosen element
if (sum % arr[i] == 0)
count++;
}
// Return the count
return count;
}
// Driver code
public static void Main()
{
int []arr = { 3, 10, 4, 6, 7 };
int n = arr.Length;
Console.WriteLine(countNum(n, arr));
}
}
// This code is contributed by inder_verma..
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// Php implementation of the approach
// Function to return the count
// of the required numbers
function countNum($N, $arr)
{
// To store the count of
// required numbers
$count = 0;
for ($i=0;$i<$N;$i++) {
// Initialize sum to 0
$Sum = 0;
for ($j = 0; $j < $N; $j++) {
// If current element and the
// chosen element are same
if ($i == $j)
continue;
// Add all other numbers of array
else
$Sum += $arr[$j];
}
// If Sum is divisible by the
// chosen element
if ($Sum % $arr[$i] == 0)
$count += 1;
}
// Return the count
return $count;
}
// Driver code
$arr = array(3, 10, 4, 6, 7);
$n = count($arr);
echo countNum($n, $arr);
// This code is contributed
// by Srathore
?>
java 描述语言
<script>
// Javascript implementation of the approach
// Function to return the count
// of the required numbers
function countNum(N, arr)
{
// To store the count of required numbers
let count = 0;
for (let i = 0; i < N; i++) {
// Initialize sum to 0
let sum = 0;
for (let j = 0; j < N; j++) {
// If current element and the
// chosen element are same
if (i == j)
continue;
// Add all other numbers of array
else
sum += arr[j];
}
// If sum is divisible by the chosen element
if (sum % arr[i] == 0)
count++;
}
// Return the count
return count;
}
let arr = [ 3, 10, 4, 6, 7 ];
let n = arr.length;
document.write(countNum(n, arr));
// This code is contributed by vaibhavrabadiya117
</script>
Output:
3
时间复杂度: O(N 2 )
高效方法:运行从 0 到 N 的单个循环，计算所有元素的总和。现在运行另一个从 0 到 N 的循环，如果(sum–arr[I])% arr[I]= 0，则递增计数。 以下是上述办法的实施情况:
C++
// C++ implementation of the approach
#include <iostream>
using namespace std;
// Function to return the count
// of the required numbers
int countNum(int N, int arr[])
{
// Initialize sum and count to 0
int sum = 0, count = 0;
// Calculate sum of all
// the array elements
for (int i = 0; i < N; i++)
sum += arr[i];
for (int i = 0; i < N; i++)
// If current element satisfies the condition
if ((sum - arr[i]) % arr[i] == 0)
count++;
// Return the count of required elements
return count;
}
// Driver code
int main()
{
int arr[] = { 3, 10, 4, 6, 7 };
int n = sizeof(arr) / sizeof(arr[0]);
cout << countNum(n, arr);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java implementation of the approach
class GFG
{
// Function to return the count
// of the required numbers
static int countNum(int N, int arr[])
{
// Initialize sum and count to 0
int sum = 0, count = 0;
// Calculate sum of all
// the array elements
for (int i = 0; i < N; i++)
{
sum += arr[i];
}
// If current element satisfies the condition
for (int i = 0; i < N; i++)
{
if ((sum - arr[i]) % arr[i] == 0)
{
count++;
}
}
// Return the count of required elements
return count;
}
// Driver code
public static void main(String[] args)
{
int arr[] = {3, 10, 4, 6, 7};
int n = arr.length;
System.out.println(countNum(n, arr));
}
}
// This code has been contributed by 29AjayKumar
Python 3
# Python3 implementation of the approach
# Function to return the count
# of the required numbers
def countNum(N, arr):
# Initialize Sum and count to 0
Sum, count = 0, 0
# Calculate Sum of all the
# array elements
for i in range(N):
Sum += arr[i]
for i in range(N):
# If current element satisfies
# the condition
if ((Sum - arr[i]) % arr[i] == 0):
count += 1
# Return the count of required
# elements
return count
# Driver code
arr = [ 3, 10, 4, 6, 7 ]
n = len(arr)
print(countNum(n, arr))
# This code is contributed
# by Mohit Kumar
C
// C# implementation of the approach
using System;
class GFG
{
// Function to return the count
// of the required numbers
static int countNum(int N, int []arr)
{
// Initialize sum and count to 0
int sum = 0, count = 0;
// Calculate sum of all
// the array elements
for (int i = 0; i < N; i++)
{
sum += arr[i];
}
// If current element satisfies the condition
for (int i = 0; i < N; i++)
{
if ((sum - arr[i]) % arr[i] == 0)
{
count++;
}
}
// Return the count of required elements
return count;
}
// Driver code
public static void Main()
{
int []arr = {3, 10, 4, 6, 7};
int n = arr.Length;
Console.WriteLine(countNum(n, arr));
}
}
/* This code contributed by PrinciRaj1992 */
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP implementation of the approach
// Function to return the count
// of the required numbers
function countNum($N, $arr)
{
// Initialize sum and count to 0
$sum = 0; $count = 0;
// Calculate sum of all
// the array elements
for ($i = 0; $i < $N; $i++)
$sum += $arr[$i];
for ($i = 0; $i < $N; $i++)
// If current element satisfies
// the condition
if (($sum - $arr[$i]) % $arr[$i] == 0)
$count++;
// Return the count of required elements
return $count;
}
// Driver code
$arr = array(3, 10, 4, 6, 7);
$n = count($arr);
echo countNum($n, $arr);
// This code contributed by Rajput-Ji
?>
java 描述语言
<script>
// Javascript implementation of the approach
// Function to return the count
// of the required numbers
function countNum(N, arr)
{
// Initialize sum and count to 0
let sum = 0, count = 0;
// Calculate sum of all
// the array elements
for (let i = 0; i < N; i++)
{
sum += arr[i];
}
// If current element satisfies the condition
for (let i = 0; i < N; i++)
{
if ((sum - arr[i]) % arr[i] == 0)
{
count++;
}
}
// Return the count of required elements
return count;
}
let arr = [3, 10, 4, 6, 7];
let n = arr.length;
document.write(countNum(n, arr));
</script>
Output:
3
时间复杂度: O(N)
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