等比例混合 n 杯后的果汁浓度
给定一个数组 arr[] ，其中 arr[i] 是IthT7】玻璃杯中果汁的浓度。任务是找出当所有玻璃以相等的比例混合时所得混合物的浓度。
示例:
输入: arr[] = {10，20，30} 输出: 20 输入: arr[] = {0，20，20} 输出: 13.3333
方法:由于果汁以相等的比例混合，因此合成浓度将是所有单个浓度的平均值。因此，需要的答案是和(arr) / n ，其中 n 是数组的大小。 以下是上述方法的实现:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// Function to return the concentration
// of the resultant mixture
double mixtureConcentration(int n, int p[])
{
double res = 0;
for (int i = 0; i < n; i++)
res += p[i];
res /= n;
return res;
}
// Driver code
int main()
{
int arr[] = { 0, 20, 20 };
int n = sizeof(arr) / sizeof(arr[0]);
cout << mixtureConcentration(n, arr);
}
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 concentration
// of the resultant mixture
static double mixtureConcentration(int n, int []p)
{
double res = 0;
for (int i = 0; i < n; i++)
res += p[i];
res /= n;
return res;
}
// Driver code
public static void main (String[] args)
{
int []arr = { 0, 20, 20 };
int n = arr.length;
System.out.println(String.format("%.4f",
mixtureConcentration(n, arr)));
}
}
// This code is contributed by chandan_jnu
Python 3
# Python3 implementation of the approach
# Function to return the concentration
# of the resultant mixture
def mixtureConcentration(n, p):
res = 0;
for i in range(n):
res += p[i];
res /= n;
return res;
# Driver code
arr = [ 0, 20, 20 ];
n = len(arr);
print(round(mixtureConcentration(n, arr), 4));
# This code is contributed
# by chandan_jnu
C
// C# implementation of the approach
using System;
class GFG
{
// Function to return the concentration
// of the resultant mixture
static double mixtureConcentration(int n, int []p)
{
double res = 0;
for (int i = 0; i < n; i++)
res += p[i];
res /= n;
return Math.Round(res,4);
}
// Driver code
static void Main()
{
int []arr = { 0, 20, 20 };
int n = arr.Length;
Console.WriteLine(mixtureConcentration(n, arr));
}
}
// This code is contributed by chandan_jnu
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP implementation of the approach
// Function to return the concentration
// of the resultant mixture
function mixtureConcentration($n, $p)
{
$res = 0;
for ($i = 0; $i < $n; $i++)
$res += $p[$i];
$res /= $n;
return $res;
}
// Driver code
$arr = array( 0, 20, 20 );
$n = count($arr);
print(round(mixtureConcentration($n, $arr), 4));
// This code is contributed by chandan_jnu
?>
java 描述语言
<script>
// Javascript implementation of the approach
// Function to return the concentration
// of the resultant mixture
function mixtureConcentration(n, p)
{
var res = 0;
for(var i = 0; i < n; i++)
res += p[i];
res /= n;
return res;
}
// Driver Code
var arr = [ 0, 20, 20 ];
var n = arr.length;
document.write(mixtureConcentration(n, arr).toFixed(4));
// This code is contributed by Ankita saini
</script>
Output:
13.3333
时间复杂度: O(N)
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