检查给定区域和斜边是否可能有直角三角形-yiteyi-C++库

给定面积和斜边，目标是打印所有边（如果可以存在直角三角形），否则打印-1。我们需要按升序打印所有面。

null

例如：

Input  : 6 5Output : 3 4 5Input  : 10 6Output : -1

推荐：请尝试你的方法 {IDE} 首先，在进入解决方案之前。

我们在下面的帖子中讨论了这个问题的解决方案。从给定的斜边和面积|集1中找出直角三角形的所有边在这篇文章中，我们将讨论一个具有以下逻辑的新解决方案。让两个未知的边为a和b 面积：A=0.5*A*b 斜边正方形：H^2=a^2+b^2 替换b，我们得到H ^2. =a ^2. +（4*A ^2. )/a ^2. 重新排列后，我们得到方程a ^4. –（H） ^2. )（a） ^2. )+4*（A） ^2. ) 这个方程的判别式D是D=H ^4. –16*（A） ^2. ) 如果D=0，则根由线性方程公式给出，根=（-b+-sqrt（D））/2*a 这些根等于边的平方，求出平方根就能得到边。

C++

                         // C++ program to check existence of                       
                         // right triangle.                       
                         #include <bits/stdc++.h>                       
                         using                                     namespace                                     std;                       
           
                         // Prints three sides of a right triangle                       
                         // from given area and hypotenuse if triangle                       
                         // is possible, else prints -1.                       
                         void                                     findRightAngle(                                     int                                     A,                                     int                                     H)                       
                         {                       
                                                 // Descriminant of the equation                       
                                                 long                                     D =                                     pow                                     (H, 4) - 16 * A * A;                       
                                   
                                                 if                                     (D >= 0)                       
                                                 {                       
                                                 // applying the linear equation                       
                                                 // formula to find both the roots                       
                                                 long                                     root1 = (H * H +                                     sqrt                                     (D)) / 2;                       
                                                 long                                     root2 = (H * H -                                     sqrt                                     (D)) / 2;                       
                                   
                                                 long                                     a =                                     sqrt                                     (root1);                       
                                                 long                                     b =                                     sqrt                                     (root2);                       
                                   
                                                 if                                     (b >= a)                       
                                                 cout << a <<                                     " "                                     << b <<                                     " "                                     << H;                       
                                                 else                       
                                                 cout << b <<                                     " "                                     << a <<                                     " "                                     << H;                       
                                                 }                       
                                                 else                       
                                                 cout <<                                     "-1"                                     ;                       
                         }                       
           
                         // Driver code                       
                         int                                     main()                       
                         {                       
                                                 findRightAngle(6, 5);                       
                                   
                         }                       
           
                         // This code is contributed By Anant Agarwal.                       

JAVA

                         // Java program to check existence of                       
                         // right triangle.                       
           
                         class                                     GFG {                       
                                   
                                                 // Prints three sides of a right triangle                       
                                                 // from given area and hypotenuse if triangle                       
                                                 // is possible, else prints -1.                       
                                                 static                                     void                                     findRightAngle(                                     double                                     A,                                     double                                     H)                       
                                                 {                       
                                                 // Descriminant of the equation                       
                                                 double                                     D = Math.pow(H,                                     4                                     ) -                                     16                                     * A * A;                       
                                   
                                                 if                                     (D >=                                     0                                     )                       
                                                 {                       
                                                 // applying the linear equation                       
                                                 // formula to find both the roots                       
                                                 double                                     root1 = (H * H + Math.sqrt(D)) /                                     2                                     ;                       
                                                 double                                     root2 = (H * H - Math.sqrt(D)) /                                     2                                     ;                       
                                   
                                                 double                                     a = Math.sqrt(root1);                       
                                                 double                                     b = Math.sqrt(root2);                       
                                                 if                                     (b >= a)                       
                                                 System.out.print(a +                                     " "                                     + b +                                     " "                                     + H);                       
                                                 else                       
                                                 System.out.print(b +                                     " "                                     + a +                                     " "                                     + H);                       
                                                 }                       
                                                 else                       
                                                 System.out.print(                                     "-1"                                     );                       
                                                 }                       
                                   
                                                 // Driver code                       
                                                 public                                     static                                     void                                     main(String arg[])                       
                                                 {                       
                                                 findRightAngle(                                     6                                     ,                                     5                                     );                       
                                                 }                       
                         }                       
           
                         // This code is contributed by Anant Agarwal.                       

Python3

                         # Python program to check existence of                       
                         # right triangle.                       
                         from                                     math                                     import                                     sqrt                       
           
                         # Prints three sides of a right triangle                       
                         # from given area and hypotenuse if triangle                       
                         # is possible, else prints -1.                       
                         def                                     findRightAngle(A, H):                       
           
                                                 # Descriminant of the equation                       
                                                 D                                     =                                     pow                                     (H,                                     4                                     )                                     -                                     16                                     *                                     A                                     *                                     A                       
                                                 if                                     D >                                     =                                     0                                     :                       
           
                                                 # applying the linear equation                       
                                                 # formula to find both the roots                       
                                                 root1                                     =                                     (H                                     *                                     H                                     +                                     sqrt(D))                                     /                                     /                                     2                       
                                                 root2                                     =                                     (H                                     *                                     H                                     -                                     sqrt(D))                                     /                                     /                                     2                       
           
                                                 a                                     =                                     int                                     (sqrt(root1))                       
                                                 b                                     =                                     int                                     (sqrt(root2))                       
                                                 if                                     b >                                     =                                     a:                       
                                                 print                                     (a, b, H)                       
                                                 else                                     :                       
                                                 print                                     (b, a, H)                       
                                                 else                                     :                       
                                                 print                                     (                                     "-1"                                     )                       
           
                         # Driver code                       
                         # Area is 6 and hypotenuse is 5.                       
                         findRightAngle(                                     6                                     ,                                     5                                     )                       

C#

                         // C# program to check existence of                       
                         // right triangle.                       
                         using                                     System;                       
           
                         class                                     GFG {                       
           
                                                 // Prints three sides of a right triangle                       
                                                 // from given area and hypotenuse if triangle                       
                                                 // is possible, else prints -1.                       
                                                 static                                     void                                     findRightAngle(                                     double                                     A,                                     double                                     H)                       
                                                 {                       
                                   
                                                 // Descriminant of the equation                       
                                                 double                                     D = Math.Pow(H, 4) - 16 * A * A;                       
           
                                                 if                                     (D >= 0) {                       
                                   
                                                 // applying the linear equation                       
                                                 // formula to find both the roots                       
                                                 double                                     root1 = (H * H + Math.Sqrt(D)) / 2;                       
                                                 double                                     root2 = (H * H - Math.Sqrt(D)) / 2;                       
           
                                                 double                                     a = Math.Sqrt(root1);                       
                                                 double                                     b = Math.Sqrt(root2);                       
                                   
                                                 if                                     (b >= a)                       
                                                 Console.WriteLine(a +                                     " "                                     + b +                                     " "                                     + H);                       
                                                 else                       
                                                 Console.WriteLine(b +                                     " "                                     + a +                                     " "                                     + H);                       
                                                 }                       
                                                 else                       
                                                 Console.WriteLine(                                     "-1"                                     );                       
                                                 }                       
           
                                                 // Driver code                       
                                                 public                                     static                                     void                                     Main()                       
                                                 {                       
                                                 findRightAngle(6, 5);                       
                                                 }                       
                         }                       
           
                         // This code is contributed by vt_m.                       

PHP

                         <?php                       
                         // PHP program to check existence of                       
                         // right triangle.                       
           
                         // Prints three sides of a right triangle                       
                         // from given area and hypotenuse if                       
                         // triangle is possible, else prints -1.                       
                         function                                     findRightAngle(                                     $A                                     ,                                     $H                                     )                       
                         {                       
                                   
                                                 // Descriminant of the equation                       
                                                 $D                                     = pow(                                     $H                                     , 4) - 16 *                                     $A                                     *                                     $A                                     ;                       
                                   
                                                 if                                     (                                     $D                                     >= 0)                       
                                                 {                       
                                   
                                                 // applying the linear equation                       
                                                 // formula to find both the roots                       
                                                 $root1                                     = (                                     $H                                     *                                     $H                                     + sqrt(                                     $D                                     )) / 2;                       
                                                 $root2                                     = (                                     $H                                     *                                     $H                                     - sqrt(                                     $D                                     )) / 2;                       
                                   
                                                 $a                                     = sqrt(                                     $root1                                     );                       
                                                 $b                                     = sqrt(                                     $root2                                     );                       
                                   
                                                 if                                     (                                     $b                                     >=                                     $a                                     )                       
                                                 echo                                     $a                                     ,                                     " "                                     ,                                     $b                                     ,                                     " "                                     ,                                     $H                                     ;                       
                                                 else                       
                                                 echo                                     $b                                     ,                                     " "                                     ,                                     $a                                     ,                                     " "                                     ,                                     $H                                     ;                       
                                                 }                       
                                                 else                       
                                                 echo                                     "-1"                                     ;                       
                         }                       
           
                                                 // Driver code                       
                                                 findRightAngle(6, 5);                       
                                   
                         // This code is contributed By Anuj_67                       
                         ?>                       

Javascript

                         <script>                       
           
                         // Javascript program to check existence of                       
                         // right triangle.                       
           
                         // Prints three sides of a right triangle                       
                         // from given area and hypotenuse if triangle                       
                         // is possible, else prints -1.                       
                         function                                     findRightAngle(A,H)                       
                         {                       
                                                 // Descriminant of the equation                       
                                                 let D = Math.pow(H, 4) - 16 * A * A;                       
                                   
                                                 if                                     (D >= 0)                       
                                                 {                       
                                                 // applying the linear equation                       
                                                 // formula to find both the roots                       
                                                 let root1 = (H * H + Math.sqrt(D)) / 2;                       
                                                 let root2 = (H * H - Math.sqrt(D)) / 2;                       
                                   
                                                 let a = Math.sqrt(root1);                       
                                                 let b = Math.sqrt(root2);                       
                                   
                                                 if                                     (b >= a)                       
                                                 document.write(a +                                     " "                                     + b +                                     " "                                     + H+                                     "<br/>"                                     );                       
                                                 else                       
                                                 document.write(b +                                     " "                                     + a +                                     " "                                     + H+                                     "<br/>"                                     );                       
                                                 }                       
                                                 else                       
                                                 document.write(                                     "-1"                                     );                       
                         }                       
           
                         // Driver code                       
           
                                                 findRightAngle(6, 5);                       
                                   
                         // This code contributed by Rajput-Ji                       
           
                         </script>                       

输出：

3 4 5

本文由 哈希特·阿格拉瓦尔 .如果你喜欢GeekSforgek，并想贡献自己的力量，你也可以使用写极客。组织或者把你的文章寄去评论-team@geeksforgeeks.org.看到你的文章出现在Geeksforgeks主页上，并帮助其他极客。如果您发现任何不正确的地方，或者您想分享有关上述主题的更多信息，请写下评论。

文章版权归作者所有，未经允许请勿转载。

THE END

技术文章