本文概述
给定一个矩形板, 长度为l, 宽度为w。我们需要将此工作表划分为正方形工作表, 以使正方形工作表的数量尽可能少。
例子:
输入:l = 4 w = 6输出:6我们可以形成边长为1个单位的正方形, 但是正方形的数目将是24, 这不是最小的。如果我们使边为2的正方形, 则我们有6个正方形。这是我们所需要的答案。同样, 我们也无法将边3设为正方形, 如果我们选择3作为正方形, 那么整张床单就无法转换成等长的糖衣。输入:l = 3 w = 5输出:15
推荐:请尝试使用{IDE}首先, 在继续解决方案之前。
正方形边的最佳长度等于GCD两个数
C ++
// CPP program to find minimum number of
// squares to make a given rectangle.
#include <bits/stdc++.h>
using namespace std;
int countRectangles( int l, int w)
{
// if we take gcd(l, w), this
// will be largest possible
// side for suare, hence minimum
// number of square.
int squareSide = __gcd(l, w);
// Number of squares.
return (l * w) / (squareSide * squareSide);
}
// Driver code
int main()
{
int l = 4, w = 6;
cout << countRectangles(l, w) << endl;
return 0;
}
Java
// Java program to find minimum number of
// squares to make a given rectangle.
class GFG{
static int __gcd( int a, int b) {
if (b== 0 ) return a;
return __gcd(b, a%b);
}
static int countRectangles( int l, int w)
{
// if we take gcd(l, w), this
// will be largest possible
// side for suare, hence minimum
// number of square.
int squareSide = __gcd(l, w);
// Number of squares.
return (l * w) / (squareSide * squareSide);
}
// Driver code
public static void main(String[] args)
{
int l = 4 , w = 6 ;
System.out.println(countRectangles(l, w));
}
}
// This code is contributed by mits
Python3
# Python3 code to find minimum number of
# squares to make a given rectangle.
import math
def countRectangles(l, w):
# if we take gcd(l, w), this
# will be largest possible
# side for suare, hence minimum
# number of square.
squareSide = math.gcd(l, w)
# Number of squares.
return (l * w) / (squareSide * squareSide)
# Driver Code
if __name__ = = '__main__' :
l = 4
w = 6
ans = countRectangles(l, w)
print ( int (ans))
# this code is contributed by
# SURENDRA_GANGWAR
C#
// C# program to find minimum number of
// squares to make a given rectangle.
class GFG{
static int __gcd( int a, int b) {
if (b==0) return a;
return __gcd(b, a%b);
}
static int countRectangles( int l, int w)
{
// if we take gcd(l, w), this
// will be largest possible
// side for suare, hence minimum
// number of square.
int squareSide = __gcd(l, w);
// Number of squares.
return (l * w) / (squareSide * squareSide);
}
// Driver code
public static void Main()
{
int l = 4, w = 6;
System.Console.WriteLine(countRectangles(l, w));
}
}
// This code is contributed by mits
的PHP
<?php
// PHP program to find minimum number
// of squares to make a given rectangle.
function gcd( $a , $b )
{
return $b ? gcd( $b , $a % $b ) : $a ;
}
function countRectangles( $l , $w )
{
// if we take gcd(l, w), this
// will be largest possible
// side for suare, hence minimum
// number of square.
$squareSide = gcd( $l , $w );
// Number of squares.
return ( $l * $w ) / ( $squareSide *
$squareSide );
}
// Driver code
$l = 4;
$w = 6;
echo countRectangles( $l , $w ) . "\n" ;
// This code is contributed
// by ChitraNayal
?>
输出如下:
6