Dados cuatro números x, y, z, n. La tarea es encontrar el número de soluciones para la ecuación x + y + z <= n, tal que 0 <= x <= X, 0 <= y <= Y, 0 <= z <= Z.
Ejemplos:
Input: x = 1, y = 1, z = 1, n = 1 Output: 4 Input: x = 1, y = 2, z = 3, n = 4 Output: 20
Enfoque: vamos a iterar explícitamente sobre todos los valores posibles de x e y (usando bucle anidado). Para uno de esos valores fijos de x e y, el problema se reduce a cuántos valores de z hay tales que z <= n – x – y y 0 <= z <= Z.
A continuación se muestra la implementación requerida para encontrar el número de soluciones:
C++
// CPP program to find the number of solutions for
// the equation x + y + z <= n, such that
// 0 <= x <= X, 0 <= y <= Y, 0 <= z <= Z.
#include <bits/stdc++.h>
using namespace std;
// function to find the number of solutions for
// the equation x + y + z <= n, such that
// 0 <= x <= X, 0 <= y <= Y, 0 <= z <= Z.
int NumberOfSolutions(int x, int y, int z, int n)
{
// to store answer
int ans = 0;
// for values of x
for (int i = 0; i <= x; i++) {
// for values of y
for (int j = 0; j <= y; j++) {
// maximum possible value of z
int temp = n - i - j;
// if z value greater than equals to 0
// then only it is valid
if (temp >= 0) {
// find minimum of temp and z
temp = min(temp, z);
ans += temp + 1;
}
}
}
// return required answer
return ans;
}
// Driver code
int main()
{
int x = 1, y = 2, z = 3, n = 4;
cout << NumberOfSolutions(x, y, z, n);
return 0;
}
C
// C program to find the number of solutions for
// the equation x + y + z <= n, such that
// 0 <= x <= X, 0 <= y <= Y, 0 <= z <= Z.
#include <stdio.h>
int min(int a, int b)
{
int min = a;
if(min > b)
min = b;
return min;
}
// function to find the number of solutions for
// the equation x + y + z <= n, such that
// 0 <= x <= X, 0 <= y <= Y, 0 <= z <= Z.
int NumberOfSolutions(int x, int y, int z, int n)
{
// to store answer
int ans = 0;
// for values of x
for (int i = 0; i <= x; i++) {
// for values of y
for (int j = 0; j <= y; j++) {
// maximum possible value of z
int temp = n - i - j;
// if z value greater than equals to 0
// then only it is valid
if (temp >= 0) {
// find minimum of temp and z
temp = min(temp, z);
ans += temp + 1;
}
}
}
// return required answer
return ans;
}
// Driver code
int main()
{
int x = 1, y = 2, z = 3, n = 4;
printf("%d",NumberOfSolutions(x, y, z, n));
return 0;
}
// This code is contributed by kothavvsaakash.
Java
// Java program to find the number of solutions for
// the equation x + y + z <= n, such that
// 0 <= x <= X, 0 <= y <= Y, 0 <= z <= Z.
import java.io.*;
class GFG {
// function to find the number of solutions for
// the equation x + y + z <= n, such that
// 0 <= x <= X, 0 <= y <= Y, 0 <= z <= Z.
static int NumberOfSolutions(int x, int y, int z, int n)
{
// to store answer
int ans = 0;
// for values of x
for (int i = 0; i <= x; i++) {
// for values of y
for (int j = 0; j <= y; j++) {
// maximum possible value of z
int temp = n - i - j;
// if z value greater than equals to 0
// then only it is valid
if (temp >= 0) {
// find minimum of temp and z
temp = Math.min(temp, z);
ans += temp + 1;
}
}
}
// return required answer
return ans;
}
// Driver code
public static void main (String[] args) {
int x = 1, y = 2, z = 3, n = 4;
System.out.println( NumberOfSolutions(x, y, z, n));
}
}
// this code is contributed by anuj_67..
Python 3
# Python3 program to find the number # of solutions for the equation # x + y + z <= n, such that # 0 <= x <= X, 0 <= y <= Y, 0 <= z <= Z. # function to find the number of solutions # for the equation x + y + z <= n, such that # 0 <= x <= X, 0 <= y <= Y, 0 <= z <= Z. def NumberOfSolutions(x, y, z, n) : # to store answer ans = 0 # for values of x for i in range(x + 1) : # for values of y for j in range(y + 1) : # maximum possible value of z temp = n - i - j # if z value greater than equals # to 0 then only it is valid if temp >= 0 : # find minimum of temp and z temp = min(temp, z) ans += temp + 1 # return required answer return ans # Driver code if __name__ == "__main__" : x, y, z, n = 1, 2, 3, 4 # function calling print(NumberOfSolutions(x, y, z, n)) # This code is contributed by ANKITRAI1
C#
// C# program to find the number of solutions for
// the equation x + y + z <= n, such that
// 0 <= x <= X, 0 <= y <= Y, 0 <= z <= Z.
using System;
public class GFG{
// function to find the number of solutions for
// the equation x + y + z <= n, such that
// 0 <= x <= X, 0 <= y <= Y, 0 <= z <= Z.
static int NumberOfSolutions(int x, int y, int z, int n)
{
// to store answer
int ans = 0;
// for values of x
for (int i = 0; i <= x; i++) {
// for values of y
for (int j = 0; j <= y; j++) {
// maximum possible value of z
int temp = n - i - j;
// if z value greater than equals to 0
// then only it is valid
if (temp >= 0) {
// find minimum of temp and z
temp = Math.Min(temp, z);
ans += temp + 1;
}
}
}
// return required answer
return ans;
}
// Driver code
static public void Main (){
int x = 1, y = 2, z = 3, n = 4;
Console.WriteLine( NumberOfSolutions(x, y, z, n));
}
}
// This code is contributed by anuj_67..
PHP
<?php
// PHP program to find the number
// of solutions for the equation
// x + y + z <= n, such that
// 0 <= x <= X, 0 <= y <= Y, 0 <= z <= Z.
// function to find the number of
// solutions for the equation
// x + y + z <= n, such that
// 0 <= x <= X, 0 <= y <= Y, 0 <= z <= Z.
function NumberOfSolutions($x, $y, $z, $n)
{
// to store answer
$ans = 0;
// for values of x
for ($i = 0; $i <= $x; $i++)
{
// for values of y
for ($j = 0; $j <= $y; $j++)
{
// maximum possible value of z
$temp = $n - $i - $j;
// if z value greater than equals
// to 0 then only it is valid
if ($temp >= 0)
{
// find minimum of temp and z
$temp = min($temp, $z);
$ans += $temp + 1;
}
}
}
// return required answer
return $ans;
}
// Driver code
$x = 1; $y = 2;
$z = 3; $n = 4;
echo NumberOfSolutions($x, $y, $z, $n);
// This code is contributed
// by Akanksha Rai(Abby_akku)
?>
Javascript
<script>
// Javascript program to find the number
// of solutions for the equation x + y + z <= n,
// such that 0 <= x <= X, 0 <= y <= Y, 0 <= z <= Z.
// Function to find the number of solutions for
// the equation x + y + z <= n, such that
// 0 <= x <= X, 0 <= y <= Y, 0 <= z <= Z.
function NumberOfSolutions(x, y, z, n)
{
// To store answer
var ans = 0;
// for values of x
for(var i = 0; i <= x; i++)
{
// for values of y
for(var j = 0; j <= y; j++)
{
// Maximum possible value of z
var temp = n - i - j;
// If z value greater than equals to 0
// then only it is valid
if (temp >= 0)
{
// Find minimum of temp and z
temp = Math.min(temp, z);
ans += temp + 1;
}
}
}
// Return required answer
return ans;
}
// Driver Code
var x = 1, y = 2, z = 3, n = 4;
document.write(NumberOfSolutions(x, y, z, n));
// This code is contributed by Ankita saini
</script>
Producción:
20
Publicación traducida automáticamente
Artículo escrito por pawan_asipu y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA