位置数，使得元素的 K 值大于所有其他元素的总和
原文:https://www . geeksforgeeks . org/position-number-to-element-add-k 大于所有其他元素的总和/
给定一个数组 arr[] 和一个数字 K 。任务是找出有效位置 i 的数量，使得 (arr[i] + K) 大于除 arr[i] 之外的数组 所有元素的和。 示例:****
Input: arr[] = {2, 1, 6, 7} K = 4
Output: 1
Explanation: There is only 1 valid position i.e 4th.
After adding 4 to the element at 4th position
it is greater than the sum of all other
elements of the array.
Input: arr[] = {2, 1, 5, 4} K = 2
Output: 0
Explanation: There is no valid position.
进场:
- 首先找到数组中所有元素的和，存储在一个变量中，比如 sum 。
- 现在，遍历数组，对于每个位置 i 检查条件(arr[I]+K)>(sum–arr[I])是否成立。
- 如果是，则增加计数器，最后打印计数器的值。
以下是上述方法的实现:
C++
// C++ program to implement above approach
#include <bits/stdc++.h>
using namespace std;
// Function that will find out
// the valid position
int validPosition(int arr[], int N, int K)
{
int count = 0, sum = 0;
// find sum of all the elements
for (int i = 0; i < N; i++) {
sum += arr[i];
}
// adding K to the element and check
// whether it is greater than sum of
// all other elements
for (int i = 0; i < N; i++) {
if ((arr[i] + K) > (sum - arr[i]))
count++;
}
return count;
}
// Driver code
int main()
{
int arr[] = { 2, 1, 6, 7 }, K = 4;
int N = sizeof(arr) / sizeof(arr[0]);
cout << validPosition(arr, N, K);
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 that will find out
// the valid position
static int validPosition(int arr[], int N, int K)
{
int count = 0, sum = 0;
// find sum of all the elements
for (int i = 0; i < N; i++)
{
sum += arr[i];
}
// adding K to the element and check
// whether it is greater than sum of
// all other elements
for (int i = 0; i < N; i++)
{
if ((arr[i] + K) > (sum - arr[i]))
count++;
}
return count;
}
// Driver code
public static void main(String[] args)
{
int arr[] = { 2, 1, 6, 7 }, K = 4;
int N = arr.length;
System.out.println(validPosition(arr, N, K));
}
}
/* This code contributed by PrinciRaj1992 */
Python 3
# Python3 program to implement
# above approach
# Function that will find out
# the valid position
def validPosition(arr, N, K):
count = 0; sum = 0;
# find sum of all the elements
for i in range(N):
sum += arr[i];
# adding K to the element and check
# whether it is greater than sum of
# all other elements
for i in range(N):
if ((arr[i] + K) > (sum - arr[i])):
count += 1;
return count;
# Driver code
arr = [2, 1, 6, 7 ];
K = 4;
N = len(arr);
print(validPosition(arr, N, K));
# This code is contributed by 29AjayKumar
C
// C# implementation of the approach
using System;
class GFG
{
// Function that will find out
// the valid position
static int validPosition(int []arr, int N, int K)
{
int count = 0, sum = 0;
// find sum of all the elements
for (int i = 0; i < N; i++)
{
sum += arr[i];
}
// adding K to the element and check
// whether it is greater than sum of
// all other elements
for (int i = 0; i < N; i++)
{
if ((arr[i] + K) > (sum - arr[i]))
count++;
}
return count;
}
// Driver code
public static void Main(String[] args)
{
int []arr = { 2, 1, 6, 7 };int K = 4;
int N = arr.Length;
Console.WriteLine(validPosition(arr, N, K));
}
}
// This code has been contributed by 29AjayKumar
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP program to implement above approach
// Function that will find out
// the valid position
function validPosition($arr, $N, $K)
{
$count = 0; $sum = 0;
// find sum of all the elements
for ($i = 0; $i < $N; $i++)
{
$sum += $arr[$i];
}
// adding K to the element and check
// whether it is greater than sum of
// all other elements
for ($i = 0; $i < $N; $i++)
{
if (($arr[$i] + $K) > ($sum - $arr[$i]))
$count++;
}
return $count;
}
// Driver code
$arr = array( 2, 1, 6, 7 );
$K = 4;
$N = count($arr) ;
echo validPosition($arr, $N, $K);
// This code is contributed by AnkitRai01
?>
java 描述语言
<script>
// Javascript program to implement above approach
// Function that will find out
// the valid position
function validPosition(arr, N, K)
{
var count = 0, sum = 0;
// find sum of all the elements
for (var i = 0; i < N; i++) {
sum += arr[i];
}
// adding K to the element and check
// whether it is greater than sum of
// all other elements
for (var i = 0; i < N; i++) {
if ((arr[i] + K) > (sum - arr[i]))
count++;
}
return count;
}
// Driver code
var arr = [ 2, 1, 6, 7 ], K = 4;
var N = arr.length;
document.write( validPosition(arr, N, K));
</script>
Output:
1
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