notepad++ 偶數(shù)行

The problem is we have a number N and we have to find sum of first N Even natural numbers.

問題是我們有一個(gè)數(shù)N ，我們必須找到前N個(gè)偶數(shù)自然數(shù)之和。

Example:

例：

Input:n = 3Output:288 (2^3 + 4^3+6^3)

A simple solution is given below...

下面給出一個(gè)簡單的解決方案 ...

Example 1:

范例1：

#include <iostream> using namespace std; int calculate(int n) { int sum = 0; for (int i = 1; i <= n; i++) sum = sum + (2*i) * (2*i) * (2*i); return sum; } int main() {int num = 3;cout<<"Number is = "<<num<<endl; cout << "Sum of cubes of first "<<num<<" even number is ="<<calculate(num); return 0; }

Output

輸出量

Number is = 3 Sum of cubes of first 3 even number is =288

The efficient approach is discussed below:

下面討論了有效的方法 ：

The sum of cubes of first n natural numbers is given by = (n*(n+1) / 2)^2 Sum of cubes of first n natural numbers can be written as... = 2^3 + 4^3 + .... + (2n)^3 Now take out common term i.e 2^3 = 2^3 * (1^3 + 2^3 + .... + n^3)= 2^3* (n*(n+1) / 2)^2= 8 * ((n^2)(n+1)^2)/4= 2 * n^2(n+1)^2

Now we can apply this formula directly to find the sum of cubes of first n even numbers.

現(xiàn)在我們可以直接應(yīng)用此公式來找到前n個(gè)偶數(shù)的立方和 。

Example 2:

范例2：

#include <iostream> using namespace std; int calculate(int n) {int sum = 2 * n * n * (n + 1) * (n + 1);return sum; } int main() {int num = 3;cout<<"Number is = "<<num<<endl; cout << "Sum of cubes of first "<<num<<" even number is ="<<calculate(num); return 0; }

Output

輸出量

Number is = 3 Sum of cubes of first 3 even number is =288

翻譯自: https://www.includehelp.com/cpp-programs/find-sum-of-cubes-of-first-n-even-numbers.aspx

notepad++ 偶數(shù)行

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