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📜  总和等于给定数N |的最小平方数设置2

📅  最后修改于: 2021-05-04 20:44:24             🧑  作者: Mango

数字始终可以表示为其他数字的平方和。注意1是一个正方形，我们总是可以将数字破译为(1 * 1 + 1 * 1 + 1 * 1 +…) 。给定数字N ，任务是将N表示为最小平方数的总和。
例子：

先决条件：总和等于给定数N的最小平方数
方法：这是动态编程的典型应用。当我们从N = 6开始时，我们可以通过减去1的平方即1乘以4，并减去2的平方即4乘以1来达到2。因此，2的子问题被调用了两次。
由于再次调用了相同的子问题，因此此问题具有“重叠子问题”属性。最小平方和问题具有动态规划问题的两个属性(请参阅此内容)。像其他典型的动态编程(DP)问题一样，可以通过自下而上的方式构造临时数组table [] []来避免相同子问题的重新计算。
下面是上述方法的实现：

C++
// C++ program to represent N as the
// sum of minimum square numbers.
#include 
using namespace std;
 
// Function for finding
// minimum square numbers
vector minSqrNum(int n)
{
  // A[i] of array arr store
  // minimum count of
  // square number to get i
  int arr[n + 1], k;
 
  // sqrNum[i] store last
  // square number to get i
  int sqrNum[n + 1];
  vector v;
 
  // Initialize
  arr[0] = 0;
  sqrNum[0] = 0;
 
  // Find minimun count of
  // square number for
  // all value 1 to n
  for (int i = 1; i <= n; i++)
  {
    // In worst case it will
    // be arr[i-1]+1 we use all
    // combination of a[i-1] and add 1
    arr[i] = arr[i - 1] + 1;
    sqrNum[i] = 1;
 
    k = 1;
    // Check for all square
    // number less or equal to i
    while (k * k <= i)
    {
      // if it gives less
      // count then update it
      if (arr[i] > arr[i - k * k] + 1)
      {
        arr[i] = arr[i - k * k] + 1;
        sqrNum[i] = k * k;
      }
      k++;
    }
  }
 
  // Vector v stores optimum
  // square number whose sum give N
  while (n > 0)
  {
    v.push_back(sqrNum[n]);
    n -= sqrNum[n];
  }
  return v;
}
 
// Driver code
int main()
{
  int n = 10;
 
  vector v;
 
  // Calling funcion
  v = minSqrNum(n);
 
  // Printing vector
  for (auto i = v.begin();
            i != v.end(); i++)
  {
    cout << *i;
    if (i + 1 != v.end())
      cout << " + ";
  }
  return 0;
}

Java
// Java program to represent
// N as the sum of minimum
// square numbers.
import java.util.*;
class GFG{
 
// Function for finding
// minimum square numbers
static Vector minSqrNum(int n)
{
  // A[i] of array arr store
  // minimum count of
  // square number to get i
  int []arr = new int[n + 1];
  int k = 0;
 
  // sqrNum[i] store last
  // square number to get i
  int []sqrNum = new int[n + 1];
  Vector v = new Vector<>();
 
  // Initialize
  arr[0] = 0;
  sqrNum[0] = 0;
 
  // Find minimun count of
  // square number for
  // all value 1 to n
  for (int i = 1; i <= n; i++)
  {
    // In worst case it will
    // be arr[i-1]+1 we use all
    // combination of a[i-1] and add 1
    arr[i] = arr[i - 1] + 1;
    sqrNum[i] = 1;
 
    k = 1;
    // Check for all square
    // number less or equal to i
    while (k * k <= i)
    {
      // if it gives less
      // count then update it
      if (arr[i] > arr[i - k * k] + 1)
      {
        arr[i] = arr[i - k * k] + 1;
        sqrNum[i] = k * k;
      }
      k++;
    }
  }
 
  // Vector v stores optimum
  // square number whose sum give N
  while (n > 0)
  {
    v.add(sqrNum[n]);
    n -= sqrNum[n];
  }
  return v;
}
 
// Driver code
public static void main(String[] args)
{
  int n = 10;
 
  Vector v;
 
  // Calling funcion
  v = minSqrNum(n);
 
  // Printing vector
  for (int i = 0; i

Python3
# Python3 program to represent N as the
# sum of minimum square numbers.
 
# Function for finding
# minimum square numbers
def minSqrNum(n):
 
    # arr[i] of array arr store
    # minimum count of
    # square number to get i
    arr = [0] * (n + 1)
     
    # sqrNum[i] store last
    # square number to get i
    sqrNum = [0] * (n + 1)
    v = []
 
    # Find minimun count of
    # square number for
    # all value 1 to n
    for i in range(n + 1):
         
        # In worst case it will
        # be arr[i-1]+1 we use all
        # combination of a[i-1] and add 1
        arr[i] = arr[i - 1] + 1
        sqrNum[i] = 1
 
        k = 1;
         
        # Check for all square
        # number less or equal to i
        while (k * k <= i):
             
            # If it gives less
            # count then update it
            if (arr[i] > arr[i - k * k] + 1):
                arr[i] = arr[i - k * k] + 1
                sqrNum[i] = k * k
 
            k += 1
 
    # v stores optimum
    # square number whose sum give N
    while (n > 0):
        v.append(sqrNum[n])
        n -= sqrNum[n];
         
    return v
 
# Driver code
n = 10
 
# Calling funcion
v = minSqrNum(n)
 
# Printing vector
for i in range(len(v)):
    print(v[i], end = "")
     
    if (i < len(v) - 1):
        print(" + ", end = "")
         
# This article is contributed by Apurvaraj

C#
// C# program to represent
// N as the sum of minimum
// square numbers.
using System;
using System.Collections.Generic;
class GFG{
 
// Function for finding
// minimum square numbers
static List minSqrNum(int n)
{
  // A[i] of array arr store
  // minimum count of
  // square number to get i
  int []arr = new int[n + 1];
  int k = 0;
 
  // sqrNum[i] store last
  // square number to get i
  int []sqrNum = new int[n + 1];
  List v = new List();
 
  // Initialize
  arr[0] = 0;
  sqrNum[0] = 0;
 
  // Find minimun count of
  // square number for
  // all value 1 to n
  for (int i = 1; i <= n; i++)
  {
    // In worst case it will
    // be arr[i-1]+1 we use all
    // combination of a[i-1] and add 1
    arr[i] = arr[i - 1] + 1;
    sqrNum[i] = 1;
 
    k = 1;
    // Check for all square
    // number less or equal to i
    while (k * k <= i)
    {
      // if it gives less
      // count then update it
      if (arr[i] > arr[i - k * k] + 1)
      {
        arr[i] = arr[i - k * k] + 1;
        sqrNum[i] = k * k;
      }
      k++;
    }
  }
 
  // List v stores optimum
  // square number whose sum give N
  while (n > 0)
  {
    v.Add(sqrNum[n]);
    n -= sqrNum[n];
  }
  return v;
}
 
// Driver code
public static void Main(String[] args)
{
  int n = 10;
 
  List v;
 
  // Calling funcion
  v = minSqrNum(n);
 
  // Printing vector
  for (int i = 0; i

输出： 
1 + 9