给定一张大小为 A x B 的纸。任务是将纸剪成任意大小的正方形。找出可以从纸上剪下的最少正方形数。
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,如果我们想从纸上切出最少数量的正方形,那么我们必须首先从纸上切出最大的正方形,并且最大的正方形将与纸的较小边具有相同的边。例如,如果纸张的尺寸为 13 x 29,那么最大的正方形将是边 13。因此我们可以切割 2 个尺寸为 13 x 13 (29/13 = 2) 的正方形。现在剩余的纸将有 3 x 13 的尺寸。类似地,我们可以使用 4 个尺寸为 3 x 3 的正方形和 3 个 1 x 1 的正方形切割剩余的纸。因此,从尺寸为 13 的纸上至少可以切割 9 个正方形× 29。
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
Javascript
输出:
9
如果您希望与专家一起参加现场课程,请参阅DSA 现场工作专业课程和学生竞争性编程现场课程。