直角等腰三角形内可容纳的最大2×2正方形数-yiteyi-C++库

在给定基底的直角等腰三角形中，2×2个单位大小的正方形的最大数量是多少（单位）。正方形的一侧必须与三角形的底部平行。

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例如：

Input  : 8
Output : 6
Please refer below diagram for explanation.

图片[1]-直角等腰三角形内可容纳的最大2×2正方形数-yiteyi-C++库

Input : 7
Output : 3

建议：请在“上解决” 实践 “首先，在讨论解决方案之前。

因为三角形是等腰的，所以给定的底也等于高度。现在在对角线部分，我们总是需要一个额外的长度2个单位的高度和三角形的基础，以适应一个三角形。（图中三角形的CF和AM段。不构成任何正方形的部分）。在基底的剩余长度中，我们可以构造长度为1/2的正方形。因为每个正方形有两个单位，所以高度的情况也是一样，不需要再计算。因此，对于给定长度的每一层，我们可以构造“（长度-2）/2”平方。这为我们提供了一个“长度-2”的基础。继续这个过程，以获得所有可用“长度2”高度的方块数，我们可以计算方块数。

while length > 2
    answer += (length - 2 )/2
    length = length - 2

为了更有效的方法，我们可以使用AP n*（n+1）/2之和的公式，其中n=长度–2

C++

                         // C++ program to count number of 2 x 2                       
                         // squares in a right isosceles triangle                       
                         #include<bits/stdc++.h>                       
                         using                                     namespace                                     std;                       
                                   
                         int                                     numberOfSquares(                                     int                                     base)                       
                         {                       
                                                 // removing the extra part we would                       
                                                 // always need                       
                                                 base = (base - 2);                       
                                   
                                                 // Since each square has base of                       
                                                 // length of 2                       
                                                 base =                                     floor                                     (base / 2);                       
                                   
                                                 return                                     base * (base + 1)/2;                       
                         }                       
                                   
                         // Driver code                       
                         int                                     main()                       
                         {                       
                                                 int                                     base = 8;                       
                                                 cout << numberOfSquares(base);                       
                                                 return                                     0;                       
                         }                       
                         // This code is improved by heroichitesh.                       

JAVA

                         // Java program to count number of 2 x 2                       
                         // squares in a right isosceles triangle                       
                                   
                         class                                     Squares                       
                         {                       
                                                 public                                     static                                     int                                     numberOfSquares(                                     int                                     base)                       
                                                 {                       
                                                 // removing the extra part                       
                                                 // we would always need                       
                                                 base = (base -                                     2                                     );                       
                                   
                                                 // Since each square has                       
                                                 // base of length of 2                       
                                                 base = Math.floorDiv(base,                                     2                                     );                       
                                   
                                                 return                                     base * (base +                                     1                                     )/                                     2                                     ;                       
                                                 }                       
                                   
                                                 // Driver code                       
                                                 public                                     static                                     void                                     main(String args[])                       
                                                 {                       
                                   
                                                 int                                     base =                                     8                                     ;                       
                                                 System.out.println(numberOfSquares(base));                       
                                                 }                       
                         }                       
                                   
                         // This code is contributed by Anshika Goyal and improved by heroichitesh.                       

Python3

                         # Python3 program to count number                       
                         # of 2 x 2 squares in a right                       
                         # isosceles triangle                       
                         def                                     numberOfSquares(base):                       
                                   
                                                 # removing the extra part we would                       
                                                 # always need                       
                                                 base                                     =                                     (base                                     -                                     2                                     )                       
                                   
                                                 # Since each square has base of                       
                                                 # length of 2                       
                                                 base                                     =                                     base                                     /                                     /                                     2                       
                                   
                                                 return                                     base                                     *                                     (base                                     +                                     1                                     )                                     /                                     2                       
                                   
                         # Driver code                       
                         base                                     =                                     8                       
                                   
                         print                                     (numberOfSquares(base))                       
                                   
                         # This code is contributed by Anant Agarwal and improved by heroichitesh.                       

C#

                         // C# program to count number of 2 x 2                       
                         // squares in a right isosceles triangle                       
                         using                                     System;                       
                                   
                         class                                     GFG {                       
                                   
                                                 public                                     static                                     int                                     numberOfSquares(                                     int                                     _base)                       
                                                 {                       
                                   
                                                 // removing the extra part                       
                                                 // we would always need                       
                                                 _base = (_base - 2);                       
                                   
                                                 // Since each square has                       
                                                 // base of length of 2                       
                                                 _base = _base / 2;                       
                                   
                                                 return                                     _base * (_base + 1)/2;                       
                                                 }                       
                                   
                                                 // Driver code                       
                                                 public                                     static                                     void                                     Main()                       
                                                 {                       
                                   
                                                 int                                     _base = 8;                       
                                                 Console.WriteLine(numberOfSquares(_base));                       
                                                 }                       
                         }                       
                                   
                         // This code is contributed by anuj_67.                       

PHP

                         <?php                       
                         // PHP program to count number of 2 x 2                       
                         // squares in a right isosceles triangle                       
                                   
                                                 function                                     numberOfSquares(                                     $base                                     )                       
                                                 {                       
                                   
                                                 // removing the extra                       
                                                 // part we would                       
                                                 // always need                       
                                                 $base                                     = (                                     $base                                     - 2);                       
                                   
                                                 // Since each square                       
                                                 // has base of                       
                                                 // length of 2                       
                                                 $base                                     = intdiv(                                     $base                                     , 2);                       
                                   
                                                 return                                     $base                                     * (                                     $base                                     + 1)/2;                       
                                                 }                       
                                   
                         // Driver code                       
                         $base                                     = 8;                       
                         echo                                     numberOfSquares(                                     $base                                     );                       
                                   
                         // This code is contributed by anuj_67 and improved by heroichitesh.                       
                         ?>                       

Javascript

                         <script>                       
                                   
                                                 // Program to count number of 2 x 2                       
                                                 // squares in a right isosceles triangle                       
                                   
                                                 function                                     numberOfSquares(base)                       
                                                 {                       
                                   
                                                 // Removing the extra part we would                       
                                                 // always need                       
                                                 base = (base - 2);                       
                                   
                                                 // Since each square has base of                       
                                                 // length of 2                       
                                                 base = Math.floor(base / 2);                       
                                   
                                                 return                                     base * (base + 1) / 2;                       
                                                 }                       
                                   
                                                 // Driver code                       
                                                 let base = 8;                       
                                                 document.write(numberOfSquares(base));                       
                                   
                                                 // This code is contributed by Mayank Tyagi                       
                                   
                         </script>                       

输出：

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