Cuente el número de triángulos posibles para el rango de lados dado

Posted on by Rudeus Greyrat

Dados cuatro números enteros A , B , C y D , la tarea es encontrar el número de conjuntos distintos (X, Y y Z) donde X, Y y Z denotan la longitud de los lados que forman un triángulo válido. A ≤ X ≤ B, B ≤ Y ≤ C y C ≤ Z ≤ D.
Ejemplos: 
 

Entrada: A = 2, B = 3, C = 4, D = 5 
Salida: 7 
Explicación: 
La longitud posible del lado de los triángulos es: 
{(2, 3, 4), (2, 4, 4), (2, 4, 5), (3, 3, 4), (3, 3, 5), (3, 4, 4) y (3, 4, 5)}
Entrada: A = 1, B = 1, C = 2 , D = 2 
Salida: 1 
Explicación: 
La única longitud posible de los lados que podemos elegir para el triángulo es (1, 2, 2) 
 

Enfoque ingenuo: la observación clave en el problema es que, si X, Y y Z son los lados válidos de un triángulo y X ≤ Y ≤ Z, entonces las condiciones suficientes para que estos lados formen un triángulo válido serán X+Y > Z .
Finalmente, el recuento del posible valor Z para X e Y dados se puede calcular como: 
 

  1. Si X+Y es mayor que D, para este caso Z se puede elegir entre [C, D], los valores totales posibles de Z serán (D-C+1).
  2. Si X+Y es menor que D y mayor que C, entonces Z puede elegirse entre [C, X+Y-1].
  3. Si X+Y es menor o igual que C, entonces no podemos elegir Z ya que estos lados no formarán un triángulo válido.

Complejidad del tiempo: O(B * C)

Enfoque eficiente: la idea es iterar para todos los valores posibles de A y luego calcular la cantidad de valores posibles de Y y Z para la X dada mediante cálculos matemáticos.
Para un X dado, el valor de X+Y estará en el rango de  [X+B, X+C] . Si calculamos el número de valores posibles mayores que D, entonces el número total de valores posibles de Y y Z será: 
 

// Number of possible values of Y and Z
// If num_greater is the number of possible
// Y values which is greater than D
Possible values = num greater * (D - C + 1)

De manera similar, sean R y L el límite superior y el límite inferior de los valores de X+Y en el rango de C y D. Entonces, las combinaciones totales para Y y Z serán: 
 

Possible Values = \frac{R*(R+1)}{2} - \frac{L*(L+1)}{2}

A continuación se muestra la implementación del enfoque anterior:
 

C++

// C++ implementation to count the
// number of possible triangles
// for the given sides ranges
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to count the number of
// possible triangles for the given
// sides ranges
int count_triangles(int a, int b,
                    int c, int d)
{
 
    int ans = 0;
 
    // Iterate for every possible of x
    for (int x = a; x <= b; ++x) {
 
        // Range of y is [b, c]
        // From this range First
        // we will find the number
        // of x + y greater than d
        int num_greater_than_d = max(d, c + x) - max(d, b + x - 1);
 
        // For x+y greater than d
        // we can choose all z from [c, d]
        // Total permutation will be
        ans += num_greater_than_d * (d - c + 1);
 
        // Now we will find the number
        // of x+y in between the [c, d]
        int r = min(max(c, c + x), d) - c;
        int l = min(max(c, b + x - 1), d) - c;
 
        // [l, r] will be the range
        // from total [c, d] x+y belongs
        // For any r such that r = x+y
        // We can choose z, in the range
        // [c, d] only less than r,
        // Thus total permutation be
        int x1 = (r * (r + 1)) / 2;
        int x2 = (l * (l + 1)) / 2;
 
        ans += x1 - x2;
    }
 
    return ans;
}
 
// Driver Code
int main()
{
    int a = 2, b = 3, c = 4, d = 5;
 
    cout << count_triangles(a, b, c, d)
         << endl;
 
    return 0;
}

Java

// Java implementation to count the
// number of possible triangles
// for the given sides ranges
import java.util.Scanner;
import java.util.Arrays;
 
class GFG{
 
// Function to count the number of
// possible triangles for the given
// sides ranges
public static int count_triangles(int a, int b,
                                  int c, int d)
{
    int ans = 0;
 
    // Iterate for every possible of x
    for(int x = a; x <= b; ++x)
    {
 
        // Range of y is [b, c]
        // From this range First
        // we will find the number
        // of x + y greater than d
        int num_greater_than_d = Math.max(d, c + x) -
                                 Math.max(d, b + x - 1);
 
        // For x+y greater than d
        // we can choose all z from [c, d]
        // Total permutation will be
        ans += num_greater_than_d * (d - c + 1);
 
        // Now we will find the number
        // of x+y in between the [c, d]
        int r = Math.min(Math.max(c, c + x), d) - c;
        int l = Math.min(Math.max(c, b + x - 1), d) - c;
 
        // [l, r] will be the range
        // from total [c, d] x+y belongs
        // For any r such that r = x+y
        // We can choose z, in the range
        // [c, d] only less than r,
        // Thus total permutation be
        int x1 = (r * (r + 1)) / 2;
        int x2 = (l * (l + 1)) / 2;
 
        ans += x1 - x2;
    }
    return ans;
}
 
// Driver Code
public static void main(String args[])
{
    int a = 2, b = 3, c = 4, d = 5;
 
    System.out.println(count_triangles(a, b, c, d));
}
}
 
// This code is contributed by SoumikMondal

Python3

# Python3 implementation to count 
# the number of possible triangles
# for the given sides ranges
 
# Function to count the number of
# possible triangles for the given
# sides ranges
def count_triangles(a, b, c, d):
 
    ans = 0
 
    # Iterate for every possible of x
    for x in range(a, b + 1):
 
        # Range of y is [b, c]
        # From this range First
        # we will find the number
        # of x + y greater than d
        num_greater_than_d = (max(d, c + x) -
                              max(d, b + x - 1))
 
        # For x+y greater than d we
        # can choose all z from [c, d]
        # Total permutation will be
        ans = (ans + num_greater_than_d *
                             (d - c + 1))
 
        # Now we will find the number
        # of x+y in between the [c, d]
        r = min(max(c, c + x), d) - c;
        l = min(max(c, b + x - 1), d) - c;
 
        # [l, r] will be the range
        # from total [c, d] x+y belongs
        # For any r such that r = x+y
        # We can choose z, in the range
        # [c, d] only less than r,
        # Thus total permutation be
        x1 = int((r * (r + 1)) / 2)
        x2 = int((l * (l + 1)) / 2)
 
        ans = ans + (x1 - x2)
 
    return ans
 
# Driver Code
a = 2
b = 3
c = 4
d = 5
 
print (count_triangles(a, b, c, d), end = '\n')
 
# This code is contributed by PratikBasu    

C#

// C# implementation to count the
// number of possible triangles
// for the given sides ranges
using System;
 
class GFG{
 
// Function to count the number of
// possible triangles for the given
// sides ranges
public static int count_triangles(int a, int b,
                                  int c, int d)
{
    int ans = 0;
 
    // Iterate for every possible of x
    for(int x = a; x <= b; ++x)
    {
 
        // Range of y is [b, c]
        // From this range First
        // we will find the number
        // of x + y greater than d
        int num_greater_than_d = Math.Max(d, c + x) -
                                 Math.Max(d, b + x - 1);
 
        // For x+y greater than d
        // we can choose all z from [c, d]
        // Total permutation will be
        ans += num_greater_than_d * (d - c + 1);
 
        // Now we will find the number
        // of x+y in between the [c, d]
        int r = Math.Min(Math.Max(c, c + x), d) - c;
        int l = Math.Min(Math.Max(c, b + x - 1), d) - c;
 
        // [l, r] will be the range
        // from total [c, d] x+y belongs
        // For any r such that r = x+y
        // We can choose z, in the range
        // [c, d] only less than r,
        // Thus total permutation be
        int x1 = (r * (r + 1)) / 2;
        int x2 = (l * (l + 1)) / 2;
 
        ans += x1 - x2;
    }
    return ans;
}
 
// Driver Code
public static void Main(String []args)
{
    int a = 2, b = 3, c = 4, d = 5;
 
    Console.WriteLine(count_triangles(a, b, c, d));
}
}
 
// This code is contributed by gauravrajput1

Javascript

<script>
 
// JavaScript implementation to count the
// number of possible triangles
// for the given sides ranges
 
// Function to count the number of
// possible triangles for the given
// sides ranges
function count_triangles(a , b, c , d)
{
    var ans = 0;
 
    // Iterate for every possible of x
    for(x = a; x <= b; ++x)
    {
 
        // Range of y is [b, c]
        // From this range First
        // we will find the number
        // of x + y greater than d
        var num_greater_than_d = Math.max(d, c + x) -
                                 Math.max(d, b + x - 1);
 
        // For x+y greater than d
        // we can choose all z from [c, d]
        // Total permutation will be
        ans += num_greater_than_d * (d - c + 1);
 
        // Now we will find the number
        // of x+y in between the [c, d]
        var r = Math.min(Math.max(c, c + x), d) - c;
        var l = Math.min(Math.max(c, b + x - 1), d) - c;
 
        // [l, r] will be the range
        // from total [c, d] x+y belongs
        // For any r such that r = x+y
        // We can choose z, in the range
        // [c, d] only less than r,
        // Thus total permutation be
        var x1 = (r * (r + 1)) / 2;
        var x2 = (l * (l + 1)) / 2;
 
        ans += x1 - x2;
    }
    return ans;
}
 
// Driver Code
 
var a = 2, b = 3, c = 4, d = 5;
 
document.write(count_triangles(a, b, c, d));
 
// This code contributed by shikhasingrajput
 
</script>
Producción: 

7

 

Tiempo Complejidad : O(ba)

Complejidad espacial: O (1), ya que no estamos usando ningún espacio adicional

Publicación traducida automáticamente

Artículo escrito por shivamsinghal1012 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA

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