剪纸成最小平方数

给定大小为A x B的纸张。任务是将纸张切成任意大小的正方形。找到可以从纸上切出的最小平方数。

例子：

Input  : 13 x 29
Output : 9
Explanation : 
2 (squares of size 13x13) + 
4 (squares of size 3x3) + 
3 (squares of size 1x1)=9

Input  : 4 x 5
Output : 5
Explanation : 
1 (squares of size 4x4) + 
4 (squares of size 1x1)

我们知道，如果要从纸上裁切最小的正方形，则必须首先从纸上裁切最大的正方形，并且最大的裁切面与纸的较小面相同。例如，如果纸张尺寸为13 x 29，则最大正方形将在面13处。因此，我们可以裁切2个尺寸为13 x 13的正方形(29/13 = 2)。现在剩余的纸张将具有3 x 13的尺寸。类似地，我们可以通过使用4个3 x 3的正方形和3个1 x 1的正方形来裁切剩余的纸张。因此，可以从13 x 29的Paper中裁切最少9个正方形。

下面是上述方法的实现。

C++

// C++ program to find minimum number of squares
// to cut a paper.
#include
using namespace std;
  
// Returns min number of squares needed
int minimumSquare(int a, int b)
{
    long long result = 0, rem = 0;
  
    // swap if a is small size side .
    if (a < b)
        swap(a, b);
  
    // Iterate until small size side is
    // greater then 0
    while (b > 0)
    {
        // Update result
        result += a/b;
  
        long long rem = a % b;
        a = b;
        b = rem;
    }
  
    return result;
}
  
// Driver code
int main()
{
    int n = 13, m = 29;
    cout << minimumSquare(n, m);
    return 0;
}

Java

// Java program to find minimum 
// number of squares to cut a paper.
class GFG{
      
// To swap two numbers
static void swap(int a,int b)
{
    int temp = a;
    a = b;
    b = temp;
}
  
// Returns min number of squares needed
static int minimumSquare(int a, int b)
{
    int result = 0, rem = 0;
  
    // swap if a is small size side .
    if (a < b)
        swap(a, b);
  
    // Iterate until small size side is
    // greater then 0
    while (b > 0)
    {
        // Update result
        result += a/b;
        rem = a % b;
        a = b;
        b = rem;
    }
  
    return result;
}
  
// Driver code
public static void main(String[] args)
{
    int n = 13, m = 29;
    System.out.println(minimumSquare(n, m));
}
}
  
//This code is contributed by Smitha Dinesh Semwal.

Python3

# Python 3 program to find minimum 
# number of squares to cut a paper.
  
# Returns min number of squares needed
def minimumSquare(a, b):
  
    result = 0
    rem = 0
  
    # swap if a is small size side .
    if (a < b):
        a, b = b, a 
  
    # Iterate until small size side is
    # greater then 0
    while (b > 0):
      
        # Update result
        result += int(a / b) 
  
        rem = int(a % b) 
        a = b 
        b = rem 
  
    return result 
  
# Driver code
n = 13
m = 29
  
print(minimumSquare(n, m))
  
# This code is contributed by
# Smitha Dinesh Semwal

C#

// C# program to find minimum 
// number of squares to cut a paper.
using System;
      
class GFG
{
      
// To swap two numbers
static void swap(int a, int b)
{
    int temp = a;
    a = b;
    b = temp;
}
  
// Returns min number of squares needed
static int minimumSquare(int a, int b)
{
    int result = 0, rem = 0;
  
    // swap if a is small size side .
    if (a < b)
        swap(a, b);
  
    // Iterate until small size side is
    // greater then 0
    while (b > 0)
    {
        // Update result
        result += a / b;
        rem = a % b;
        a = b;
        b = rem;
    }
    return result;
}
  
// Driver code
public static void Main(String[] args)
{
    int n = 13, m = 29;
    Console.WriteLine(minimumSquare(n, m));
}
}
  
// This code is contributed by 29AjayKumar

输出：

9

请注意，上述贪婪解决方案并不总是能产生最佳结果。例如，如果输入为36 x 30，则上述算法将产生输出6，但是我们可以将纸张切成5个正方形
1)三个大小为12 x 12的正方形
2)大小为18 x 18的两个正方形。

感谢Sergey V. Pereslavtsev指出了上述情况。