总和等于给定数N |的最小平方数套装3

先决条件：拉格朗日的四平方定理，勒让德的三平方定理

数字始终可以表示为其他数字的平方和。请注意，1是一个正方形，我们总是可以将数字打破为(1 * 1 + 1 * 1 + 1 * 1 +…)。给定一个数字n，求和为N的最小平方数。

例子：

Input: N = 13 
Output: 2 
Explanation: 
13 can be expressed as, 13 = 3² + 2². Hence, the output is 2.

Input: N = 100 
Output: 1 
Explanation: 
100 can be expressed as 100 = 10². Hence, the output is 1.

天真的方法：有关[O(N * sqrt(N))]方法，请参阅本文的Set 2 。

高效方法：要优化幼稚方法，该想法是使用拉格朗日的四平方定理和勒让德的三平方定理。下面讨论这两个定理：

Lagrange’s four-square theorem, also known as Bachet’s conjecture, states that every natural number can be represented as the sum of four integer squares, where each integer is non-negative. 

Legendre’s three-square theorem states that a natural number can be represented as the sum of three squares of integers if and only if n is not of the form : n = 4^a (8b+7), for non-negative integers a and b. 

因此，证明了代表任何数字N的最小平方数只能在集合{1,2,3,4}内。因此，仅检查这四个可能的值，就可以找到代表任何数字N的最小平方数。请按照以下步骤操作：

• 如果N是一个完美的平方，则结果为1 。
• 如果N可以表示为两个平方的和，则结果为2 。
• 如果N不能以N = 4 ^a (8b + 7)的形式表示，其中a和b是非负整数，则根据勒让德三平方定理，结果为3 。
• 如果不满足上述所有条件，则根据拉格朗日的四平方定理，结果为4 。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
  
// Function that returns true if N
// is a perfect square
bool isPerfectSquare(int N)
{
    int floorSqrt = sqrt(N);
  
    return (N == floorSqrt * floorSqrt);
}
  
// Function that returns true check if
// N is sum of three squares
bool legendreFunction(int N)
{
    // Factor out the powers of 4
    while (N % 4 == 0)
        N /= 4;
  
    // N is NOT of the
    // form 4^a * (8b + 7)
    if (N % 8 != 7)
        return true;
    else
        return false;
}
  
// Function that finds the minimum
// number of square whose sum is N
int minSquares(int N)
{
  
    // If N is perfect square
    if (isPerfectSquare(N))
        return 1;
  
    // If N is sum of 2 perfect squares
    for (int i = 1; i * i < N; i++) {
        if (isPerfectSquare(N - i * i))
            return 2;
    }
  
    // If N is sum of 3 perfect squares
    if (legendreFunction(N))
        return 3;
  
    // Otherwise, N is the
    // sum of 4 perfect squares
    return 4;
}
  
// Driver code
int main()
{
    // Given number
    int N = 123;
  
    // Function call
    cout << minSquares(N);
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
  
class GFG{
  
// Function that returns true if N
// is a perfect square
static boolean isPerfectSquare(int N)
{
    int floorSqrt = (int)Math.sqrt(N);
  
    return (N == floorSqrt * floorSqrt);
}
  
// Function that returns true check if
// N is sum of three squares
static boolean legendreFunction(int N)
{
      
    // Factor out the powers of 4
    while (N % 4 == 0)
        N /= 4;
  
    // N is NOT of the
    // form 4^a * (8b + 7)
    if (N % 8 != 7)
        return true;
    else
        return false;
}
  
// Function that finds the minimum
// number of square whose sum is N
static int minSquares(int N)
{
  
    // If N is perfect square
    if (isPerfectSquare(N))
        return 1;
  
    // If N is sum of 2 perfect squares
    for(int i = 1; i * i < N; i++) 
    {
        if (isPerfectSquare(N - i * i))
            return 2;
    }
  
    // If N is sum of 3 perfect squares
    if (legendreFunction(N))
        return 3;
  
    // Otherwise, N is the
    // sum of 4 perfect squares
    return 4;
}
  
// Driver code
public static void main(String[] args)
{
      
    // Given number
    int N = 123;
  
    // Function call
    System.out.print(minSquares(N));
}
}
  
// This code is contributed by 29AjayKumar

Python3

# Python3 program for the above approach
from math import sqrt, floor, ceil
  
# Function that returns True if N
# is a perfect square
def isPerfectSquare(N):
    
    floorSqrt = floor(sqrt(N))
    return (N == floorSqrt * floorSqrt)
    
# Function that returns True check if
# N is sum of three squares
def legendreFunction(N):
    
    # Factor out the powers of 4
    while (N % 4 == 0):
        N //= 4
  
    # N is NOT of the
    # form 4^a * (8b + 7)
    if (N % 8 != 7):
        return True
    else:
        return False
  
# Function that finds the minimum
# number of square whose sum is N
def minSquares(N):
    
    # If N is perfect square
    if (isPerfectSquare(N)):
        return 1
  
    # If N is sum of 2 perfect squares
    for i in range(N):
        if i * i < N:
            break
        if (isPerfectSquare(N - i * i)):
            return 2
            
    # If N is sum of 3 perfect squares
    if (legendreFunction(N)):
        return 3
        
    # Otherwise, N is the
    # sum of 4 perfect squares
    return 4
    
# Driver code
if __name__ == '__main__':
  
    # Given number
    N = 123
  
    # Function call
    print(minSquares(N))
  
# This code is contributed by mohit kumar 29

C#

// C# program for the above approach
using System;
  
class GFG{
  
// Function that returns true if N
// is a perfect square
static bool isPerfectSquare(int N)
{
    int floorSqrt = (int)Math.Sqrt(N);
  
    return (N == floorSqrt * floorSqrt);
}
  
// Function that returns true check 
// if N is sum of three squares
static bool legendreFunction(int N)
{
      
    // Factor out the powers of 4
    while (N % 4 == 0)
        N /= 4;
  
    // N is NOT of the
    // form 4^a * (8b + 7)
    if (N % 8 != 7)
        return true;
    else
        return false;
}
  
// Function that finds the minimum
// number of square whose sum is N
static int minSquares(int N)
{
  
    // If N is perfect square
    if (isPerfectSquare(N))
        return 1;
  
    // If N is sum of 2 perfect squares
    for(int i = 1; i * i < N; i++) 
    {
        if (isPerfectSquare(N - i * i))
            return 2;
    }
  
    // If N is sum of 3 perfect squares
    if (legendreFunction(N))
        return 3;
  
    // Otherwise, N is the
    // sum of 4 perfect squares
    return 4;
}
  
// Driver code
public static void Main(String[] args)
{
      
    // Given number
    int N = 123;
  
    // Function call
    Console.Write(minSquares(N));
}
}
  
// This code is contributed by Rajput-Ji

输出：

3

时间复杂度： O(sqrt(N))
辅助空间： O(1)