给定一个正整数 N .任务是找到1 2. + 2 2. + 3 2. + ….. + N 2. .
null
例如：
Input : N = 4Output : 3012 + 22 + 32 + 42= 1 + 4 + 9 + 16= 30Input : N = 5Output : 55
方法1:O（N） 这个想法是从1到n运行一个循环，对于每个i，1<=i<=n，找到i 2. 总而言之。
下面是这种方法的实现
C++
// CPP Program to find sum of square of first n natural numbers #include <bits/stdc++.h> using namespace std; // Return the sum of square of first n natural numbers int squaresum( int n) { // Iterate i from 1 and n // finding square of i and add to sum. int sum = 0; for ( int i = 1; i <= n; i++) sum += (i * i); return sum; } // Driven Program int main() { int n = 4; cout << squaresum(n) << endl; return 0; } |
JAVA
// Java Program to find sum of // square of first n natural numbers import java.io.*; class GFG { // Return the sum of square of first n natural numbers static int squaresum( int n) { // Iterate i from 1 and n // finding square of i and add to sum. int sum = 0 ; for ( int i = 1 ; i <= n; i++) sum += (i * i); return sum; } // Driven Program public static void main(String args[]) throws IOException { int n = 4 ; System.out.println(squaresum(n)); } } /*This code is contributed by Nikita Tiwari.*/ |
Python3
# Python3 Program to # find sum of square # of first n natural # numbers # Return the sum of # square of first n # natural numbers def squaresum(n) : # Iterate i from 1 # and n finding # square of i and # add to sum. sm = 0 for i in range ( 1 , n + 1 ) : sm = sm + (i * i) return sm # Driven Program n = 4 print (squaresum(n)) # This code is contributed by Nikita Tiwari.*/ |
C#
// C# Program to find sum of // square of first n natural numbers using System; class GFG { // Return the sum of square of first // n natural numbers static int squaresum( int n) { // Iterate i from 1 and n // finding square of i and add to sum. int sum = 0; for ( int i = 1; i <= n; i++) sum += (i * i); return sum; } // Driven Program public static void Main() { int n = 4; Console.WriteLine(squaresum(n)); } } /* This code is contributed by vt_m.*/ |
PHP
<?php // PHP Program to find sum of // square of first n natural numbers // Return the sum of square of // first n natural numbers function squaresum( $n ) { // Iterate i from 1 and n // finding square of i and // add to sum. $sum = 0; for ( $i = 1; $i <= $n ; $i ++) $sum += ( $i * $i ); return $sum ; } // Driven Code $n = 4; echo (squaresum( $n )); // This code is contributed by Ajit. ?> |
Javascript
<script> // Javascript Program to find sum of square of first n natural numbers // Return the sum of square of first n natural numbers function squaresum(n) { // Iterate i from 1 and n // finding square of i and add to sum. let sum = 0; for (let i = 1; i <= n; i++) sum += (i * i); return sum; } // Driven Program let n = 4; document.write(squaresum(n) + "<br>" ); // This code is contributed by Mayank Tyagi </script> |
输出：
30
方法2:O（1）
前N个自然数的平方和=（N*（N+1）*（2*N+1））/6
例如 对于N=4，求和=（4*（4+1）*（2*4+1））/6 = 180 / 6 = 30 对于N=5，求和=（5*（5+1）*（2*5+1））/6 = 55
证据：
We know,(k + 1)3 = k3 + 3 * k2 + 3 * k + 1We can write the above identity for k from 1 to n:23 = 13 + 3 * 12 + 3 * 1 + 1 ......... (1)33 = 23 + 3 * 22 + 3 * 2 + 1 ......... (2)43 = 33 + 3 * 32 + 3 * 3 + 1 ......... (3)53 = 43 + 3 * 42 + 3 * 4 + 1 ......... (4)...n3 = (n - 1)3 + 3 * (n - 1)2 + 3 * (n - 1) + 1 ......... (n - 1)(n + 1)3 = n3 + 3 * n2 + 3 * n + 1 ......... (n)Putting equation (n - 1) in equation n,(n + 1)3 = (n - 1)3 + 3 * (n - 1)2 + 3 * (n - 1) + 1 + 3 * n2 + 3 * n + 1 = (n - 1)3 + 3 * (n2 + (n - 1)2) + 3 * ( n + (n - 1) ) + 1 + 1By putting all equation, we get(n + 1)3 = 13 + 3 * Σ k2 + 3 * Σ k + Σ 1n3 + 3 * n2 + 3 * n + 1 = 1 + 3 * Σ k2 + 3 * (n * (n + 1))/2 + nn3 + 3 * n2 + 3 * n = 3 * Σ k2 + 3 * (n * (n + 1))/2 + nn3 + 3 * n2 + 2 * n - 3 * (n * (n + 1))/2 = 3 * Σ k2n * (n2 + 3 * n + 2) - 3 * (n * (n + 1))/2 = 3 * Σ k2n * (n + 1) * (n + 2) - 3 * (n * (n + 1))/2 = 3 * Σ k2n * (n + 1) * (n + 2 - 3/2) = 3 * Σ k2n * (n + 1) * (2 * n + 1)/2 = 3 * Σ k2n * (n + 1) * (2 * n + 1)/6 = Σ k2
以下是该方法的实施情况：
C++
// CPP Program to find sum // of square of first n // natural numbers #include <bits/stdc++.h> using namespace std; // Return the sum of square of // first n natural numbers int squaresum( int n) { return (n * (n + 1) * (2 * n + 1)) / 6; } // Driven Program int main() { int n = 4; cout << squaresum(n) << endl; return 0; } |
JAVA
// Java Program to find sum // of square of first n // natural numbers import java.io.*; class GFG { // Return the sum of square // of first n natural numbers static int squaresum( int n) { return (n * (n + 1 ) * ( 2 * n + 1 )) / 6 ; } // Driven Program public static void main(String args[]) throws IOException { int n = 4 ; System.out.println(squaresum(n)); } } /*This code si contributed by Nikita Tiwari.*/ |
Python3
# Python3 Program to # find sum of square # of first n natural # numbers # Return the sum of # square of first n # natural numbers def squaresum(n) : return (n * (n + 1 ) * ( 2 * n + 1 )) / / 6 # Driven Program n = 4 print (squaresum(n)) #This code is contributed by Nikita Tiwari. |
C#
// C# Program to find sum // of square of first n // natural numbers using System; class GFG { // Return the sum of square // of first n natural numbers static int squaresum( int n) { return (n * (n + 1) * (2 * n + 1)) / 6; } // Driven Program public static void Main() { int n = 4; Console.WriteLine(squaresum(n)); } } /*This code is contributed by vt_m.*/ |
PHP
<?php // PHP Program to find sum // of square of first n // natural numbers // Return the sum of square of // first n natural numbers function squaresum( $n ) { return ( $n * ( $n + 1) * (2 * $n + 1)) / 6; } // Driven Code $n = 4; echo (squaresum( $n )); // This code is contributed by Ajit. ?> |
Javascript
<script> // Javascript program to find sum // of square of first n // natural numbers // Return the sum of square of // first n natural numbers function squaresum(n) { return parseInt((n * (n + 1) * (2 * n + 1)) / 6); } // Driver code let n = 4; document.write(squaresum(n)); // This code is contributed by rishavmahato348 </script> |
输出：
30
避免提前溢出： 对于较大的n，（n*（n+1）*（2*n+1））的值将溢出。利用n*（n+1）必须能被2整除的事实，我们可以在一定程度上避免溢出。
C++
// CPP Program to find sum of square of first // n natural numbers. This program avoids // overflow upto some extent for large value // of n. #include <bits/stdc++.h> using namespace std; // Return the sum of square of first n natural // numbers int squaresum( int n) { return (n * (n + 1) / 2) * (2 * n + 1) / 3; } // Driven Program int main() { int n = 4; cout << squaresum(n) << endl; return 0; } |
Python3
# Python Program to find sum of square of first # n natural numbers. This program avoids # overflow upto some extent for large value # of n.y def squaresum(n): return (n * (n + 1 ) / 2 ) * ( 2 * n + 1 ) / 3 # main() n = 4 print (squaresum(n)); # Code Contributed by Mohit Gupta_OMG <(0_o)> |
JAVA
// Java Program to find sum of square of first // n natural numbers. This program avoids // overflow upto some extent for large value // of n. import java.io.*; import java.util.*; class GFG { // Return the sum of square of first n natural // numbers public static int squaresum( int n) { return (n * (n + 1 ) / 2 ) * ( 2 * n + 1 ) / 3 ; } public static void main (String[] args) { int n = 4 ; System.out.println(squaresum(n)); } } // Code Contributed by Mohit Gupta_OMG <(0_o)> |
C#
// C# Program to find sum of square of first // n natural numbers. This program avoids // overflow upto some extent for large value // of n. using System; class GFG { // Return the sum of square of // first n natural numbers public static int squaresum( int n) { return (n * (n + 1) / 2) * (2 * n + 1) / 3; } // Driver Code public static void Main() { int n = 4; Console.WriteLine(squaresum(n)); } } // This Code is Contributed by vt_m.> |
PHP
<?php // PHP Program to find // sum of square of first // n natural numbers. // This program avoids // overflow upto some // extent for large value // of n. // Return the sum of square // of first n natural numbers function squaresum( $n ) { return ( $n * ( $n + 1) / 2) * (2 * $n + 1) / 3; } // Driver Code $n = 4; echo squaresum( $n ) ; // This code is contributed by vt_m. ?> |
Javascript
<script> // javascript Program to find sum of square of first // n natural numbers. This program avoids // overflow upto some extent for large value // of n. // Return the sum of square of first n natural // numbers function squaresum( n) { return (n * (n + 1) / 2) * (2 * n + 1) / 3; } // Driven Program let n = 4; document.write(squaresum(n)); // This code contributed by aashish1995 </script> |
输出：
30
© 版权声明
文章版权归作者所有，未经允许请勿转载。
THE END
![关于”PostgreSQL错误：关系[表]不存在“问题的原因和解决方案-yiteyi-C++库](https://www.yiteyi.com/wp-content/themes/zibll/img/thumbnail.svg)
