打印可以相加形成给定总和的元素
原文:https://www . geesforgeks . org/print-elements-可添加到给定金额中的元素/
给定一个正整数数组 arr[] 和一个和，任务是打印将要包含的元素以获得给定的和。 注:
- 考虑队列形式的元素，即从开始到元素总和小于或等于给定总和的元素。
- 此外，数组元素的和不一定等于给定的和。
因为任务是检查柠檬是否可以包含在内。
示例:
输入: arr[] = {3，5，3，2，1}，Sum = 10 输出: 3 5 2 通过加 3，5 和 3，Sum 变成 11，所以去掉最后的 3。 然后加 2，和变成 10。所以不需要添加其他元素 。
输入: arr[] = {7，10，6，4}，总和= 12 输出: 7 4 As，7+10 和 7+6 的总和大于 12 ，但 7+4 = 11 小于 12。 所以，7 和 4 可以包含在内
进场:
- 检查添加当前元素时，总和是否小于给定的总和。
- 如果是，则添加它。
- 否则转到下一个元素，重复相同的操作，直到它们的总和小于或等于给定的总和。
下面是上述方法的实现:
C++
// C++ implementation of above approach
#include <bits/stdc++.h>
using namespace std;
// Function that finds whether an element
// will be included or not
void includeElement(int a[], int n, int sum)
{
for (int i = 0; i < n; i++) {
// Check if the current element
// will be incuded or not
if ((sum - a[i]) >= 0) {
sum = sum - a[i];
cout << a[i]<< " ";
}
}
}
// Driver Code
int main()
{
int arr[] = { 3, 5, 3, 2, 1 };
int n = sizeof(arr) / sizeof(arr[0]);
int sum = 10;
includeElement(arr, n, sum);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java implementation
// of above approach
class GFG
{
// Function that finds whether
// an element will be included
// or not
static void includeElement(int a[],
int n, int sum)
{
for (int i = 0; i < n; i++)
{
// Check if the current element
// will be included or not
if ((sum - a[i]) >= 0)
{
sum = sum - a[i];
System.out.print(a[i] + " ");
}
}
}
// Driver code
public static void main(String[] args)
{
int arr[] = { 3, 5, 3, 2, 1 };
int n = arr.length;
int sum = 10;
includeElement(arr, n, sum);
}
}
// This code is contributed by Bilal
Python 3
# Python 3 implementation of above approach
# Function that finds whether an element
# will be included or not
def includeElement(a, n, sum) :
for i in range(n) :
# Check if the current element
# will be incuded or not
if sum - a[i] >= 0 :
sum = sum - a[i]
print(a[i],end = " ")
# Driver code
if __name__ == "__main__" :
arr = [ 3, 5, 3, 2, 1]
n = len(arr)
sum = 10
includeElement(arr, n, sum)
# This code is contributed by ANKITRAI1
C
// C# implementation
// of above approach
using System;
class GFG
{
// Function that finds whether
// an element will be included
// or not
static void includeElement(int[] a,
int n, int sum)
{
for (int i = 0; i < n; i++)
{
// Check if the current element
// will be included or not
if ((sum - a[i]) >= 0)
{
sum = sum - a[i];
Console.Write(a[i] + " ");
}
}
}
// Driver code
static void Main()
{
int[] arr = new int[]{ 3, 5, 3, 2, 1 };
int n = arr.Length;
int sum = 10;
includeElement(arr, n, sum);
}
}
// This code is contributed by mits
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP implementation of above approach
// Function that finds whether an
// element will be included or not
function includeElement(&$a, $n, $sum)
{
for ($i = 0; $i < $n; $i++)
{
// Check if the current element
// will be incuded or not
if (($sum - $a[$i]) >= 0)
{
$sum = $sum - $a[$i];
echo $a[$i] . " ";
}
}
}
// Driver Code
$arr = array( 3, 5, 3, 2, 1 );
$n = sizeof($arr);
$sum = 10;
includeElement($arr, $n, $sum);
// This code is contributed
// by ChitraNayal
?>
java 描述语言
<script>
// Javascript implementation of above approach
// Function that finds whether an element
// will be included or not
function includeElement(a, n, sum)
{
for(var i = 0; i < n; i++)
{
// Check if the current element
// will be incuded or not
if ((sum - a[i]) >= 0)
{
sum = sum - a[i];
document.write( a[i] + " ");
}
}
}
// Driver Code
var arr = [ 3, 5, 3, 2, 1 ];
var n = arr.length;
var sum = 10;
includeElement(arr, n, sum);
// This code is contributed by itsok
</script>
Output:
3 5 2
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