Dados tres enteros, A, X y n, la tarea es imprimir los términos de la siguiente serie de expresiones binomiales.
(A+X) n = n C 0 A n X 0 + n C 1 A n-1 X 1 + n C 2 A n-2 X 2 +….+ n C n A 0 X n
Ejemplos:
Input : A = 1, X = 1, n = 5 Output : 1 5 10 10 5 1 Input : A = 1, B = 2, n = 6 Output : 1 12 60 160 240 192 64
Solución simple: sabemos que para cada valor de n habrá (n+1) término en la serie binomial. Así que ahora usamos un enfoque simple y calculamos el valor de cada elemento de la serie y lo imprimimos.
nCr = (n!) / ((n-r)! * (r)!) Below is value of general term. Tr+1 = nCn-rAn-rXr So at each position we have to find the value of the general term and print that term .
C++
// CPP program to print terms of binomial
// series and also calculate sum of series.
#include <bits/stdc++.h>
using namespace std;
// function to calculate factorial of
// a number
int factorial(int n)
{
int f = 1;
for (int i = 2; i <= n; i++)
f *= i;
return f;
}
// function to print the series
void series(int A, int X, int n)
{
// calculating the value of n!
int nFact = factorial(n);
// loop to display the series
for (int i = 0; i < n + 1; i++) {
// For calculating the
// value of nCr
int niFact = factorial(n - i);
int iFact = factorial(i);
// calculating the value of
// A to the power k and X to
// the power k
int aPow = pow(A, n - i);
int xPow = pow(X, i);
// display the series
cout << (nFact * aPow * xPow) /
(niFact * iFact) << " ";
}
}
// main function started
int main()
{
int A = 3, X = 4, n = 5;
series(A, X, n);
return 0;
}
Java
// Java program to print terms of binomial
// series and also calculate sum of series.
import java.io.*;
class GFG {
// function to calculate factorial of
// a number
static int factorial(int n)
{
int f = 1;
for (int i = 2; i <= n; i++)
f *= i;
return f;
}
// function to print the series
static void series(int A, int X, int n)
{
// calculating the value of n!
int nFact = factorial(n);
// loop to display the series
for (int i = 0; i < n + 1; i++) {
// For calculating the
// value of nCr
int niFact = factorial(n - i);
int iFact = factorial(i);
// calculating the value of
// A to the power k and X to
// the power k
int aPow = (int)Math.pow(A, n - i);
int xPow = (int)Math.pow(X, i);
// display the series
System.out.print((nFact * aPow * xPow)
/ (niFact * iFact) + " ");
}
}
// main function started
public static void main(String[] args)
{
int A = 3, X = 4, n = 5;
series(A, X, n);
}
}
// This code is contributed by vt_m.
Python3
# Python3 program to print terms of binomial # series and also calculate sum of series. # function to calculate factorial # of a number def factorial(n): f = 1 for i in range(2, n+1): f *= i return f # Function to print the series def series(A, X, n): # calculating the value of n! nFact = factorial(n) # loop to display the series for i in range(0, n + 1): # For calculating the # value of nCr niFact = factorial(n - i) iFact = factorial(i) # calculating the value of # A to the power k and X to # the power k aPow = pow(A, n - i) xPow = pow(X, i) # display the series print (int((nFact * aPow * xPow) / (niFact * iFact)), end = " ") # Driver Code A = 3; X = 4; n = 5 series(A, X, n) # This code is contributed by Smitha Dinesh Semwal.
C#
// C# program to print terms of binomial
// series and also calculate sum of series.
using System;
class GFG {
// function to calculate factorial of
// a number
static int factorial(int n)
{
int f = 1;
for (int i = 2; i <= n; i++)
f *= i;
return f;
}
// function to print the series
static void series(int A, int X, int n)
{
// calculating the value of n!
int nFact = factorial(n);
// loop to display the series
for (int i = 0; i < n + 1; i++) {
// For calculating the
// value of nCr
int niFact = factorial(n - i);
int iFact = factorial(i);
// calculating the value of
// A to the power k and X to
// the power k
int aPow = (int)Math.Pow(A, n - i);
int xPow = (int)Math.Pow(X, i);
// display the series
Console.Write((nFact * aPow * xPow)
/ (niFact * iFact) + " ");
}
}
// main function started
public static void Main()
{
int A = 3, X = 4, n = 5;
series(A, X, n);
}
}
// This code is contributed by anuj_67.
PHP
<?php
// PHP program to print
// terms of binomial
// series and also
// calculate sum of series.
// function to calculate
// factorial of a number
function factorial($n)
{
$f = 1;
for ($i = 2; $i <= $n; $i++)
$f *= $i;
return $f;
}
// function to print the series
function series($A, $X, $n)
{
// calculating the
// value of n!
$nFact = factorial($n);
// loop to display
// the series
for ($i = 0; $i < $n + 1; $i++)
{
// For calculating the
// value of nCr
$niFact = factorial($n - $i);
$iFact = factorial($i);
// calculating the value of
// A to the power k and X to
// the power k
$aPow = pow($A, $n - $i);
$xPow = pow($X, $i);
// display the series
echo ($nFact * $aPow * $xPow) /
($niFact * $iFact) , " ";
}
}
// Driver Code
$A = 3;
$X = 4;
$n = 5;
series($A, $X, $n);
// This code is contributed by anuj_67.
?>
Javascript
<script>
// JavaScript program to print terms of binomial
// series and also calculate sum of series.
// function to calculate factorial of
// a number
function factorial(n)
{
let f = 1;
for (let i = 2; i <= n; i++)
f *= i;
return f;
}
// function to print the series
function series(A, X, n)
{
// calculating the value of n!
let nFact = factorial(n);
// loop to display the series
for (let i = 0; i < n + 1; i++) {
// For calculating the
// value of nCr
let niFact = factorial(n - i);
let iFact = factorial(i);
// calculating the value of
// A to the power k and X to
// the power k
let aPow = Math.pow(A, n - i);
let xPow = Math.pow(X, i);
// display the series
document.write((nFact * aPow * xPow)
/ (niFact * iFact) + " ");
}
}
// Driver Code
let A = 3, X = 4, n = 5;
series(A, X, n);
// This code is contributed by chinmoy1997pal.
</script>
Producción:
243 1620 4320 5760 3840 1024
Complejidad temporal : O(n 2 )
Espacio auxiliar : O(1)
Solución eficiente:
la idea es calcular el siguiente término usando el término anterior. Podemos calcular el siguiente término en tiempo O(1). Usamos la siguiente propiedad de los coeficientes binomiales .
n C i+1 = n C i *(ni)/(i+1)
C++
// CPP program to print terms of binomial
// series and also calculate sum of series.
#include <bits/stdc++.h>
using namespace std;
// function to print the series
void series(int A, int X, int n)
{
// Calculating and printing first term
int term = pow(A, n);
cout << term << " ";
// Computing and printing remaining terms
for (int i = 1; i <= n; i++) {
// Find current term using previous terms
// We increment power of X by 1, decrement
// power of A by 1 and compute nCi using
// previous term by multiplying previous
// term with (n - i + 1)/i
term = term * X * (n - i + 1)/(i * A);
cout << term << " ";
}
}
// main function started
int main()
{
int A = 3, X = 4, n = 5;
series(A, X, n);
return 0;
}
Java
// Java program to print terms of binomial
// series and also calculate sum of series.
import java.io.*;
class GFG {
// function to print the series
static void series(int A, int X, int n)
{
// Calculating and printing first
// term
int term = (int)Math.pow(A, n);
System.out.print(term + " ");
// Computing and printing
// remaining terms
for (int i = 1; i <= n; i++) {
// Find current term using
// previous terms We increment
// power of X by 1, decrement
// power of A by 1 and compute
// nCi using previous term by
// multiplying previous term
// with (n - i + 1)/i
term = term * X * (n - i + 1)
/ (i * A);
System.out.print(term + " ");
}
}
// main function started
public static void main(String[] args)
{
int A = 3, X = 4, n = 5;
series(A, X, n);
}
}
// This code is contributed by vt_m.
Python3
# Python 3 program to print terms of binomial # series and also calculate sum of series. # Function to print the series def series(A, X, n): # Calculating and printing first term term = pow(A, n) print(term, end = " ") # Computing and printing remaining terms for i in range(1, n+1): # Find current term using previous terms # We increment power of X by 1, decrement # power of A by 1 and compute nCi using # previous term by multiplying previous # term with (n - i + 1)/i term = int(term * X * (n - i + 1)/(i * A)) print(term, end = " ") # Driver Code A = 3; X = 4; n = 5 series(A, X, n) # This code is contributed by Smitha Dinesh Semwal.
C#
// C# program to print terms of binomial
// series and also calculate sum of series.
using System;
public class GFG {
// function to print the series
static void series(int A, int X, int n)
{
// Calculating and printing first
// term
int term = (int)Math.Pow(A, n);
Console.Write(term + " ");
// Computing and printing
// remaining terms
for (int i = 1; i <= n; i++) {
// Find current term using
// previous terms We increment
// power of X by 1, decrement
// power of A by 1 and compute
// nCi using previous term by
// multiplying previous term
// with (n - i + 1)/i
term = term * X * (n - i + 1)
/ (i * A);
Console.Write(term + " ");
}
}
// main function started
public static void Main()
{
int A = 3, X = 4, n = 5;
series(A, X, n);
}
}
// This code is contributed by anuj_67.
PHP
<?php
// PHP program to print
// terms of binomial
// series and also
// calculate sum of
// series.
// function to print
// the series
function series($A, $X, $n)
{
// Calculating and printing
// first term
$term = pow($A, $n);
echo $term , " ";
// Computing and printing
// remaining terms
for ($i = 1; $i <= $n; $i++)
{
// Find current term
// using previous terms
// We increment power
// of X by 1, decrement
// power of A by 1 and
// compute nCi using
// previous term by
// multiplying previous
// term with (n - i + 1)/i
$term = $term * $X * ($n - $i + 1) /
($i * $A);
echo $term , " ";
}
}
// Driver Code
$A = 3;
$X = 4;
$n = 5;
series($A, $X, $n);
// This code is contributed by anuj_67.
?>
Javascript
<script>
// JavaScript program to print terms of binomial
// series and also calculate sum of series.
// function to print the series
function series(A, X, n)
{
// Calculating and printing first term
let term = Math.pow(A, n);
document.write(term + " ");
// Computing and printing remaining terms
for (let i = 1; i <= n; i++) {
// Find current term using previous terms
// We increment power of X by 1, decrement
// power of A by 1 and compute nCi using
// previous term by multiplying previous
// term with (n - i + 1)/i
term = term * X * (n - i + 1)/(i * A);
document.write(term + " ");
}
}
// main function started
let A = 3, X = 4, n = 5;
series(A, X, n);
// This code is contributed by Surbhi Tyagi.
</script>
Producción:
243 1620 4320 5760 3840 1024
Complejidad temporal : O(n)
Espacio auxiliar : O(1)
Publicación traducida automáticamente
Artículo escrito por Manish_100 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA