给定正整数N。任务是找到1 2 + 2 2 + 3 2 +….. + N 2 。
例子 :
Input : N = 4
Output : 30
12 + 22 + 32 + 42
= 1 + 4 + 9 + 16
= 30
Iput : N = 5
Output : 55
方法1:O(N)的想法是运行一个从1到n的循环,对于每个i,1 <= i <= n,求和2 。
下面是这种方法的实现
C++
// CPP Program to find sum of square of first n natural numbers
#include
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
Javascript
C++
// CPP Program to find sum
// of square of first n
// natural numbers
#include
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
Javascript
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
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
Javascript
输出 :
30
方法2:O(1)
Sum of squares of first N natural numbers = (N*(N+1)*(2*N+1))/6
例如
对于N = 4,总和=(4 *(4 + 1)*(2 * 4 + 1))/ 6
= 180/6
= 30
对于N = 5,求和=(5 *(5 + 1)*(2 * 5 +1))/ 6
= 55
证明:
We know,
(k + 1)3 = k3 + 3 * k2 + 3 * k + 1
We 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 + 1
By putting all equation, we get
(n + 1)3 = 13 + 3 * Σ k2 + 3 * Σ k + Σ 1
n3 + 3 * n2 + 3 * n + 1 = 1 + 3 * Σ k2 + 3 * (n * (n + 1))/2 + n
n3 + 3 * n2 + 3 * n = 3 * Σ k2 + 3 * (n * (n + 1))/2 + n
n3 + 3 * n2 + 2 * n - 3 * (n * (n + 1))/2 = 3 * Σ k2
n * (n2 + 3 * n + 2) - 3 * (n * (n + 1))/2 = 3 * Σ k2
n * (n + 1) * (n + 2) - 3 * (n * (n + 1))/2 = 3 * Σ k2
n * (n + 1) * (n + 2 - 3/2) = 3 * Σ k2
n * (n + 1) * (2 * n + 1)/2 = 3 * Σ k2
n * (n + 1) * (2 * n + 1)/6 = Σ k2
以下是此方法的实现:
C++
// CPP Program to find sum
// of square of first n
// natural numbers
#include
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.*/
的PHP
Java脚本
输出 :
30
避免过早溢出:
对于大n,(n *(n + 1)*(2 * n + 1))的值将溢出。我们可以使用n *(n + 1)必须被2整除的事实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
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.>
的PHP
Java脚本
输出:
30