当arr [i] = i *(-1)^ i时数组中从索引L到R的元素的总和

给定两个整数$L$和$R$还有一个数组arr [] ，其中每个元素都位于索引处$i$计算为arr [i] = i *(-1) ⁱ 。任务是在索引范围内找到数组的这些元素的总和$[L, R]$ 。

例子：

Input: L = 1 , R = 5 
Output: -3 
Sum = (-1) + 2 + (-3) + 4 + (-5) = -3

Input: L = 5 , R = 100000000 
Output: 49999998 

天真的方法：根据数组元素的定义，数组的每个奇数元素为负，偶数元素为正。因此，要找到总和，请执行一个从(L到R)的for循环，并保持所有奇数(负)数和偶数(正数)的和。最后，返回总和。

有效方法：可以注意到，该系列所有奇数元素的总和将始终等于(totalOdd) ² ，其中totalOdd =奇数元素的总数，偶数之和为totalEven *(totalEven + 1) 。现在，我们要做的就是找到直到L和R为止的所有奇数元素的总和。存储两者之差以获得L和R之间所有奇数元素的总和。对偶数进行同样的操作，最后返回偶数和奇数和的差。

下面是上述方法的实现：

C++

// C++ implementation of above approach
#include 
using namespace std;
 
// function to return the odd sum
long int Odd_Sum(int n)
{
 
    // total odd elements upto n
    long int total = (n + 1) / 2;
 
    // sum of odd elements upto n
    long int odd = total * total;
 
    return odd;
}
 
// function to return the even sum
long int Even_Sum(int n)
{
 
    // total even elements upto n
    long int total = (n) / 2;
 
    // sum of even elements upto n
    long int even = total * (total + 1);
 
    return even;
}
 
// Function to find sum from L to R.
int sumLtoR(int L, int R)
{
 
    long int odd_sum, even_sum;
 
    odd_sum = Odd_Sum(R) - Odd_Sum(L - 1);
 
    even_sum = Even_Sum(R) - Even_Sum(L - 1);
 
    // return final sum from L to R
    return even_sum - odd_sum;
}
 
// Driver Program
int main()
{
 
    int L = 1, R = 5;
 
    // function call to print answer
    cout << sumLtoR(L, R);
 
    return 0;
}

Java

// Java implementation of above approach
 
import java.io.*;
 
class GFG {
     
 
 
// function to return the odd sum
static long  Odd_Sum(int n)
{
 
    // total odd elements upto n
    long  total = (n + 1) / 2;
 
    // sum of odd elements upto n
    long  odd = total * total;
 
    return odd;
}
 
// function to return the even sum
static long  Even_Sum(int n)
{
 
    // total even elements upto n
    long  total = (n) / 2;
 
    // sum of even elements upto n
    long  even = total * (total + 1);
 
    return even;
}
 
// Function to find sum from L to R.
static long sumLtoR(int L, int R)
{
 
    long  odd_sum, even_sum;
 
    odd_sum = Odd_Sum(R) - Odd_Sum(L - 1);
 
    even_sum = Even_Sum(R) - Even_Sum(L - 1);
 
    // return final sum from L to R
    return even_sum - odd_sum;
}
 
// Driver Program
 
    public static void main (String[] args) {
        int L = 1, R = 5;
 
    // function call to print answer
    System.out.println( sumLtoR(L, R));
    }
}
// This code is contributed by shs..

Python3

# Python3 implementation of above approach
 
# function to return the odd sum
def Odd_Sum(n):
 
    # total odd elements upto n
    total =(n+1)//2
 
    # sum of odd elements upto n
    odd = total*total
    return odd
 
# function to return the even sum
def Even_Sum(n):
 
    # total even elements upto n
    total = n//2
 
    # sum of even elements upto n
    even = total*(total+1)
    return even
 
def sumLtoR(L,R):
    odd_sum = Odd_Sum(R)-Odd_Sum(L-1)
    even_sum = Even_Sum(R)- Even_Sum(L-1)
 
    # return final sum from L to R
    return even_sum-odd_sum
 
 
# Driver code
L =1; R = 5
print(sumLtoR(L,R))
 
# This code is contributed by Shrikant13

C#

// C# implementation of above approach
class GFG
{
     
// function to return the odd sum
static long Odd_Sum(int n)
{
 
    // total odd elements upto n
    long total = (n + 1) / 2;
 
    // sum of odd elements upto n
    long odd = total * total;
 
    return odd;
}
 
// function to return the even sum
static long Even_Sum(int n)
{
 
    // total even elements upto n
    long total = (n) / 2;
 
    // sum of even elements upto n
    long even = total * (total + 1);
 
    return even;
}
 
// Function to find sum from L to R.
static long sumLtoR(int L, int R)
{
    long odd_sum, even_sum;
 
    odd_sum = Odd_Sum(R) - Odd_Sum(L - 1);
 
    even_sum = Even_Sum(R) - Even_Sum(L - 1);
 
    // return final sum from L to R
    return even_sum - odd_sum;
}
 
// Driver Code
public static void Main ()
{
    int L = 1, R = 5;
 
    // function call to print answer
    System.Console.WriteLine(sumLtoR(L, R));
}
}
 
// This code is contributed by mits

PHP

> 1;
 
    // sum of odd elements upto n
    $odd = $total * $total;
 
    return $odd;
}
 
// function to return the even sum
function Even_Sum($n)
{
 
    // for total even elements upto n
    // divide by 2
    $total = $n >> 1;
     
    // sum of even elements upto n
    $even = $total * ($total + 1);
 
    return $even;
}
 
// Function to find sum from L to R.
function sumLtoR($L, $R)
{
    $odd_sum = Odd_Sum($R) -
               Odd_Sum($L - 1);
 
    $even_sum = Even_Sum($R) -
                Even_Sum($L - 1);
 
     
    // print final sum from L to R
    return $even_sum - $odd_sum ;
}
 
// Driver Code
$L = 1 ;
$R = 5;
 
// function call to print answer
echo sumLtoR($L, $R);
 
// This code is contributed by ANKITRAI1
?>

Javascript

 
// Javascript implementation of above approach
 
// Function to return the odd sum
function Odd_Sum(n)
{
     
    // Total odd elements upto n
    var total = parseInt((n + 1) / 2);
 
    // Sum of odd elements upto n
    var odd = total * total;
 
    return odd;
}
 
// Function to return the even sum
function Even_Sum(n)
{
     
    // Total even elements upto n
    var total = parseInt((n) / 2);
 
    // Sum of even elements upto n
    var even = total * (total + 1);
 
    return even;
}
 
// Function to find sum from L to R.
function sumLtoR(L, R)
{
    var odd_sum, even_sum;
 
    odd_sum = Odd_Sum(R) - Odd_Sum(L - 1);
    even_sum = Even_Sum(R) - Even_Sum(L - 1);
 
    // Return final sum from L to R
    return even_sum - odd_sum;
}
 
// Driver code
var L = 1, R = 5;
 
// Function call to print answer
document.write(sumLtoR(L, R));
 
// This code is contributed by SoumikMondal
 

输出：

-3