平方金字塔数(平方和)

方形金字塔数表示第一自然数的平方和。前几个Square金字塔数是1、5、14、30、55、91、140、204、285、385、506，…
在几何上，这些数字表示要堆叠以形成具有方形底角的金字塔的球的数量。请查看此Wiki图像以获取更多清晰度。
给定数字s(1 <= s <= 1000000000)。如果s是前n个自然数的平方和，则打印n，否则打印-1。
例子 ：

Input : 14
Output : 3
Explanation : 1*1 + 2*2 + 3*3 = 14

Input : 26
Output : -1

一个简单的解决方案是遍历所有从1开始的数字，计算当前总和。如果当前总和等于给定总和，则返回true，否则返回false。

C++

// C++ program to check if a
// given number is sum of
// squares of natural numbers.
#include 
using namespace std;
 
// Function to find if the
// given number is sum of
// the squares of first n
// natural numbers
int findS(int s)
{
    int sum = 0;
 
    // Start adding squares of
    // the numbers from 1
    for (int n = 1; sum < s; n++)
    {
        sum += n * n;
 
        // If sum becomes equal to s
        // return n
        if (sum == s)
            return n;
    }
 
    return -1;
}
 
// Drivers code
int main()
{
    int s = 13;
    int n = findS(s);
    n == -1 ? cout << "-1" : cout << n;
 
    return 0;
}

Java

// Java program to check if a
// given number is sum of
// squares of natural numbers.
class GFG
{
 
    // Function to find if the
    // given number is sum of
    // the squares of first
    // n natural numbers
    static int findS(int s)
    {
        int sum = 0;
 
        // Start adding squares of
        // the numbers from 1
        for (int n = 1; sum < s; n++)
        {
            sum += n * n;
 
            // If sum becomes equal to s
            // return n
            if (sum == s)
                return n;
        }
 
        return -1;
    }
 
    // Drivers code
    public static void main(String[] args)
    {
 
        int s = 13;
        int n = findS(s);
        if (n == -1)
            System.out.println("-1");
        else
            System.out.println(n);
    }
}

Python3

# Python3 program to find if
# the given number is sum of
# the squares of first
# n natural numbers
 
# Function to find if the given
# number is sum of the squares
# of first n natural numbers
def findS (s):
    _sum = 0
    n = 1
     
    # Start adding squares of
    # the numbers from 1
    while(_sum < s):
        _sum += n * n
        n+= 1
    n-= 1
     
    # If sum becomes equal to s
    # return n
    if _sum == s:
        return n
    return -1
 
# Driver code
s = 13
n = findS (s)
if n == -1:
    print("-1")
else:
    print(n)

C#

// C# program to check if a given
// number is sum of squares of
// natural numbers.
using System;
 
class GFG
{
     
    // Function to find if the given
    // number is sum of the squares
    // of first n natural numbers
    static int findS(int s)
    {
        int sum = 0;
     
        // Start adding squares of
        // the numbers from 1
        for (int n = 1; sum < s; n++)
        {
            sum += n * n;
     
            // If sum becomes equal
            // to s return n
            if (sum == s)
                return n;
        }
     
        return -1;
    }
     
    // Drivers code
    public static void Main()
    {
        int s = 13;
         
        int n = findS(s);
         
        if(n == -1)
            Console.Write("-1") ;
        else
            Console.Write(n);
    }
}
// This code is contribute by
// Smitha Dinesh Semwal

Javascript

 
// JavaScript to check if a
// given number is sum of
// squares of natural numbers.
 
    // Function to find if the
    // given number is sum of
    // the squares of first
    // n natural numbers
    function findS(s)
    {
        let sum = 0;
   
        // Start adding squares of
        // the numbers from 1
        for (let n = 1; sum < s; n++)
        {
            sum += n * n;
   
            // If sum becomes equal to s
            // return n
            if (sum == s)
                return n;
        }
   
        return -1;
    }
 
// Driver code
        let s = 13;
        let n = findS(s);
        if (n == -1)
            document.write("-1");
        else
            document.write(n);
 
// This code is contributed by souravghosh0416.

输出 ：

-1

另一种解决方案是使用牛顿拉夫森法。
我们知道，前n个自然数的平方和为n *(n + 1)*(2 * n +1)/ 6 。
我们可以写解决方案为
k *(k + 1)*(2 * k +1)/ 6 = s
k *(k + 1)*(2 * k +1)– 6s = 0
我们可以使用牛顿拉夫森法找到上述三次方程的根，然后检查根是否为整数。