打印数组中偶数和奇数元素之和的程序

先决条件 - 阵列基础
给定一个数组，编写一个程序，分别求偶数和奇数索引位置的值之和。
例子：

Input : arr = {1, 2, 3, 4, 5, 6}
Output :Even index positions sum 9
        Odd index positions sum 12
Explanation: Here, n = 6 so there will be 3 even 
index positions and 3 odd index positions in an array
Even = 1 + 3 + 5 = 9
Odd =  2 + 4 + 6 = 12

Input : arr = {10, 20, 30, 40, 50, 60, 70}
Output : Even index positions sum 160
        Odd index positions sum 170
Explanation: Here, n = 7 so there will be 3 odd
index positions and 4 even index positions in an array
Even = 10 + 30 + 50 + 70 = 160
Odd = 20 + 40 + 60 = 120

C++

// CPP program to find out
// Sum of elements at even and
// odd index positions separately
#include 
   
using namespace std;
   
// Function to calculate sum
void EvenOddSum(int arr[], int n)
{
    int even = 0;
    int odd = 0;
    for (int i = 0; i < n; i++) {
        // Loop to find even, odd sum
        if (i % 2 == 0)
            even += arr[i];
        else
            odd += arr[i];
    }
   
    cout << "Even index positions sum " << even;
    cout << "\nOdd index positions sum " << odd;
}
   
// Driver function
int main()
{
    int arr[] = { 1, 2, 3, 4, 5, 6 };
    int n = sizeof(arr) / sizeof(arr[0]);
   
    EvenOddSum(arr, n);
   
    return 0;
}

Java

// Java program to find out
// Sum of elements at even and
// odd index positions separately
import java.io.*;
 
class EvenOddSum {
    public static void main(String args[])
    {
        int arr[] = { 1, 2, 3, 4, 5, 6 };
        int even = 0, odd = 0;
 
        // Loop to find even, odd sum
        for (int i = 0; i < arr.length; i++) {
            if (i % 2 == 0)
                even += arr[i];
            else
                odd += arr[i];
        }
 
        System.out.println("Even index positions sum: " + even);
        System.out.println("Odd index positions sum: " + odd);
    }
}

Python3

# Python program to find out
# Sum of elements at even and
# odd index positions separately
 
# Function to calculate Sum
def EvenOddSum(a, n):
    even = 0
    odd = 0
    for i in range(n):
 
        # Loop to find even, odd Sum
        if i % 2 == 0:
            even += a[i]
        else:
            odd += a[i]
     
    print ("Even index positions sum ", even)
    print ("nOdd index positions sum ", odd)
 
# Driver Function
 
arr = [1, 2, 3, 4, 5, 6]
n = len(arr)
 
EvenOddSum(arr, n)
 
# This code is contributed by Sachin Bisht

C#

// C# program to find out
// Sum of elements at even and
// odd index positions separately
using System;
 
public class GFG {
     
    public static void Main()
    {
        int[] arr = { 1, 2, 3, 4, 5, 6 };
        int even = 0, odd = 0;
 
        // Loop to find even, odd sum
        for (int i = 0; i < arr.Length; i++)
        {
            if (i % 2 == 0)
                even += arr[i];
            else
                odd += arr[i];
        }
 
        Console.WriteLine("Even index positions"
                             + " sum: " + even);
                              
        Console.WriteLine("Odd index positions "
                               + "sum: " + odd);
    }
}
 
// This code is contributed by Sam007.

PHP

Javascript

Python3

# Python program to find out
# Sum of elements at even and
# odd index positions separately
 
# Function to calculate Sum
 
 
def EvenOddSum(a, n):
    even_sum = sum(a[::2])
    odd_sum = sum(a[1::2])
    print("Even index positions sum", even_sum)
    print("Odd index positions sum", odd_sum)
 
 
# Driver Function
 
arr = [1, 2, 3, 4, 5, 6]
n = len(arr)
 
EvenOddSum(arr, n)
 
# This code is contributed by vikkycirus

Javascript

输出

Even index positions sum 9
Odd index positions sum 12

方法2：在Python中使用切片：

使用切片计算所有偶数索引的总和，并对奇数索引重复相同的操作并打印总和。

下面是实现：

Python3

# Python program to find out
# Sum of elements at even and
# odd index positions separately
 
# Function to calculate Sum
 
 
def EvenOddSum(a, n):
    even_sum = sum(a[::2])
    odd_sum = sum(a[1::2])
    print("Even index positions sum", even_sum)
    print("Odd index positions sum", odd_sum)
 
 
# Driver Function
 
arr = [1, 2, 3, 4, 5, 6]
n = len(arr)
 
EvenOddSum(arr, n)
 
# This code is contributed by vikkycirus

Javascript

输出

Even index positions sum 9
Odd index positions sum 12