Dado un número entero N , la tarea es imprimir los primeros N términos de la serie en su forma de fracción, es decir
1/4, 1/2, 3/4, 1, 5/4, …
La serie anterior tiene valores como 0.25, 0.5, 0.75, 1, 1.25,….etc. Es una progresión aritmética que comienza en 0,25 y tiene una diferencia de 0,25.
Ejemplos:
Entrada: N = 6
Salida: 1/4 1/2 3/4 1 5/4 3/2
Entrada: N = 9
Salida: 1/4 1/2 3/4 1 5/4 3/2 7/4 2 9/4
Enfoque: Considere los primeros cuatro términos de la serie como términos base. Almacene los elementos del numerador y los elementos del denominador por separado.
Considere el primer término 1/4 , el quinto término es 1 + (1 * 4) / 4 que es 1/5 .
Del mismo modo, considere el segundo término 1/2 , el sexto término es 1 + (1 * 2) / 2, que es 3/2 .
Por lo tanto, podemos considerar que los denominadores siempre serán 2 , 4 o ningún denominador y el numerador del término se puede calcular a partir del denominador.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// Function to print the required series
void printSeries(int n)
{
// Numerators for the first four numerators
// of the series
int nmtr[4] = { 1, 1, 1, 3 };
// Denominators for the first four denominators
// of the series
int dntr[4] = { 0, 4, 2, 4 };
for (int i = 1; i <= n; i++) {
// If location of the term in the series is
// a multiple of 4 then there will be no denominator
if (i % 4 == 0)
cout << nmtr[i % 4] + (i / 4) - 1 << " ";
// Otherwise there will be denominator
else {
// Printing the numerator and the denominator terms
cout << nmtr[i % 4] + ((i / 4) * dntr[i % 4])
<< "/" << dntr[i % 4] << " ";
}
}
}
// Driver code
int main()
{
int n = 9;
printSeries(n);
return 0;
}
Java
// Java implementation of the approach
class GFG
{
// Function to print the required series
public static void printSeries(int n)
{
// Numerators for the first four numerators
// of the series
int[] nmtr = new int[]{ 1, 1, 1, 3 };
// Denominators for the first four denominators
// of the series
int[] dntr = new int[]{ 0, 4, 2, 4 };
for (int i = 1; i <= n; i++)
{
// If location of the term in the series is
// a multiple of 4 then there will be no denominator
if (i % 4 == 0)
System.out.print( nmtr[i % 4] + (i / 4) - 1 + " ");
// Otherwise there will be denominator
else
{
// Printing the numerator and the denominator terms
System.out.print( nmtr[i % 4] + ((i / 4) * dntr[i % 4])
+"/" + dntr[i % 4] +" ");
}
}
}
// Driver code
public static void main(String[] args)
{
int n = 9;
printSeries(n);
}
}
// This code is contributed
// by 29AjayKumar
Python3
# Python 3 implementation of the approach
# Function to print the required series
def printSeries(n):
# Numerators for the first four
# numerators of the series
nmtr = [1, 1, 1, 3]
# Denominators for the first four
# denominators of the series
dntr = [0, 4, 2, 4]
for i in range(1, n + 1, 1):
# If location of the term in the
# series is a multiple of 4 then
# there will be no denominator
if (i % 4 == 0):
print(nmtr[i % 4] + int(i / 4) - 1,
end = " ")
# Otherwise there will be denominator
else:
# Printing the numerator and
# the denominator terms
print(nmtr[i % 4] + (int(i / 4) *
dntr[i % 4]), end = "")
print("/", end = "")
print(dntr[i % 4], end = " ")
# Driver code
if __name__ == '__main__':
n = 9
printSeries(n)
# This code is contributed by
# Shashank_Sharma
C#
// C# implementation of the approach
using System;
class GFG
{
// Function to print the required series
static void printSeries(int n)
{
// Numerators for the first four numerators
// of the series
int[] nmtr = { 1, 1, 1, 3 };
// Denominators for the first four denominators
// of the series
int[] dntr = { 0, 4, 2, 4 };
for (int i = 1; i <= n; i++)
{
// If location of the term in the series is
// a multiple of 4 then there will be no denominator
if (i % 4 == 0)
Console.Write((nmtr[i % 4] + (i / 4) - 1) + " ");
// Otherwise there will be denominator
else
{
// Printing the numerator and the denominator terms
Console.Write((nmtr[i % 4] + ((i / 4) * dntr[i % 4])) +
"/" + dntr[i % 4] + " ");
}
}
}
// Driver code
public static void Main()
{
int n = 9;
printSeries(n);
}
}
// This code is contributed
// by Akanksha Rai
Javascript
<script>
// javascript implementation of the approach
// Function to print the required series
function printSeries( n)
{
// Numerators for the first four numerators
// of the series
let nmtr = [ 1, 1, 1, 3 ];
// Denominators for the first four denominators
// of the series
let dntr = [ 0, 4, 2, 4 ];
for (let i = 1; i <= n; i++)
{
// If location of the term in the series is
// a multiple of 4 then there will be no denominator
if (i % 4 == 0)
document.write( nmtr[i % 4] + (i / 4) - 1 + " ");
// Otherwise there will be denominator
else {
// Printing the numerator and the denominator terms
document.write( nmtr[i % 4] + (parseInt(i / 4) * dntr[i % 4])
+ "/" + dntr[i % 4] + " ");
}
}
}
// Driver code
let n = 9;
printSeries(n);
// This code is contributed by Rajput-Ji
</script>
Producción:
1/4 1/2 3/4 1 5/4 3/2 7/4 2 9/4