Dado un entero positivo D y una string M que representa el día y el mes de un año bisiesto, la tarea es encontrar la fecha posterior al próximo medio año.
Ejemplos:
Entrada: D = 15, M = “Enero”
Salida: 16 de julio
Explicación: La fecha del 15 de enero al próximo medio año es el 16 de julio.Entrada: D = 10, M = “Octubre”
Salida: 10 de abril
Enfoque: dado que un año bisiesto contiene 366 días , el problema dado se puede resolver encontrando los datos después de incrementar la fecha actual en 183 días . Siga los pasos para resolver el problema:
- Almacene la cantidad de días de cada mes en esa array, digamos days[] .
- Inicialice una variable, digamos curMonth como M , para almacenar el índice del mes actual.
- Inicialice una variable, diga curDate como D , para almacenar la fecha actual.
- Inicialice una variable, digamos contar como 183 , que representa la cuenta de días para incrementar.
- Iterar hasta que el valor de count sea positivo y realizar los siguientes pasos:
- Si el valor de count es positivo y curDate es como máximo el número de días en curMonth , disminuya el valor de count en 1 e incremente el valor de curDate en 1 .
- Si el valor de count es 0 , salga del bucle .
- Actualice el valor de curMonth por (curMonth + 1)%12 .
- Después de completar los pasos anteriores, imprima la fecha correspondiente a curDate y curMonth como resultado.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ program for the above approach
#include <iostream>
using namespace std;
// Function to find the date
// after the next half-year
void getDate(int d, string m) {
// Stores the number of days in the
// months of a leap year
int days[] = {31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
// List of months
string month[] = {"January", "February",
"March", "April",
"May", "June",
"July", "August",
"September", "October",
"November", "December"};
// Days in half of a year
int cnt = 183;
// Index of current month
int cur_month;
for(int i = 0; i < 12; i++)
if(m == month[i])
cur_month = i;
// Starting day
int cur_date = d;
while(1) {
while(cnt > 0 && cur_date <= days[cur_month]) {
// Decrement the value of
// cnt by 1
cnt -= 1;
// Increment cur_date
cur_date += 1;
}
// If cnt is equal to 0, then
// break out of the loop
if(cnt == 0)
break;
// Update cur_month
cur_month = (cur_month + 1) % 12;
// Update cur_date
cur_date = 1;
}
// Print the resultant date
cout << cur_date << " " << month[cur_month] << endl;
}
// Driver Code
int main() {
int D = 15;
string M = "January";
// Function Call
getDate(D, M);
return 0;
}
// This code is contributed by Dharanendra L V.
Java
// Java program for the above approach
class GFG{
// Function to find the date
// after the next half-year
public static void getDate(int d, String m)
{
// Stores the number of days in the
// months of a leap year
int[] days = { 31, 29, 31, 30, 31, 30,
31, 31, 30, 31, 30, 31 };
// List of months
String[] month = { "January", "February", "March",
"April", "May", "June", "July",
"August", "September", "October",
"November", "December" };
// Days in half of a year
int cnt = 183;
// Index of current month
int cur_month = 0;
for(int i = 0; i < 12; i++)
if (m == month[i])
cur_month = i;
// Starting day
int cur_date = d;
while (true)
{
while (cnt > 0 && cur_date <= days[cur_month])
{
// Decrement the value of
// cnt by 1
cnt -= 1;
// Increment cur_date
cur_date += 1;
}
// If cnt is equal to 0, then
// break out of the loop
if (cnt == 0)
break;
// Update cur_month
cur_month = (cur_month + 1) % 12;
// Update cur_date
cur_date = 1;
}
// Print the resultant date
System.out.println(cur_date + " " +
month[cur_month]);
}
// Driver Code
public static void main(String args[])
{
int D = 15;
String M = "January";
// Function Call
getDate(D, M);
}
}
// This code is contributed by SoumikMondal
Python3
# Python program for the above approach # Function to find the date # after the next half-year def getDate(d, m): # Stores the number of days in the # months of a leap year days = [31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31] # List of months month = ['January', 'February', 'March', 'April', 'May', 'June', 'July', 'August', 'September', 'October', 'November', 'December'] # Days in half of a year cnt = 183 # Index of current month cur_month = month.index(m) # Starting day cur_date = d while(1): while(cnt > 0 and cur_date <= days[cur_month]): # Decrement the value of # cnt by 1 cnt -= 1 # Increment cur_date cur_date += 1 # If cnt is equal to 0, then # break out of the loop if(cnt == 0): break # Update cur_month cur_month = (cur_month + 1) % 12 # Update cur_date cur_date = 1 # Print the resultant date print(cur_date, month[cur_month]) # Driver Code D = 15 M = "January" # Function Call getDate(D, M)
C#
// C# program for the above approach
using System;
class GFG{
// Function to find the date
// after the next half-year
static void getDate(int d, string m)
{
// Stores the number of days in the
// months of a leap year
int[] days = { 31, 29, 31, 30, 31, 30,
31, 31, 30, 31, 30, 31 };
// List of months
string[] month = { "January", "February", "March",
"April", "May", "June", "July",
"August", "September", "October",
"November", "December" };
// Days in half of a year
int cnt = 183;
// Index of current month
int cur_month = 0;
for(int i = 0; i < 12; i++)
if (m == month[i])
cur_month = i;
// Starting day
int cur_date = d;
while (true)
{
while (cnt > 0 && cur_date <= days[cur_month])
{
// Decrement the value of
// cnt by 1
cnt -= 1;
// Increment cur_date
cur_date += 1;
}
// If cnt is equal to 0, then
// break out of the loop
if (cnt == 0)
break;
// Update cur_month
cur_month = (cur_month + 1) % 12;
// Update cur_date
cur_date = 1;
}
// Print the resultant date
Console.WriteLine(cur_date + " " +
month[cur_month]);
}
// Driver Code
public static void Main()
{
int D = 15;
string M = "January";
// Function Call
getDate(D, M);
}
}
// This code is contributed by ukasp
Javascript
<script>
// Javascript program for the above approach
// Function to find the date
// after the next half-year
function getDate(d, m)
{
// Stores the number of days in the
// months of a leap year
let days = [ 31, 29, 31, 30, 31, 30,
31, 31, 30, 31, 30, 31 ];
// List of months
let month = [ "January", "February", "March",
"April", "May", "June", "July",
"August", "September", "October",
"November", "December" ];
// Days in half of a year
let cnt = 183;
// Index of current month
let cur_month = 0;
for(let i = 0; i < 12; i++)
if (m == month[i])
cur_month = i;
// Starting day
let cur_date = d;
while (true)
{
while (cnt > 0 && cur_date <= days[cur_month])
{
// Decrement the value of
// cnt by 1
cnt -= 1;
// Increment cur_date
cur_date += 1;
}
// If cnt is equal to 0, then
// break out of the loop
if (cnt == 0)
break;
// Update cur_month
cur_month = (cur_month + 1) % 12;
// Update cur_date
cur_date = 1;
}
// Print the resultant date
document.write(cur_date + " " +
month[cur_month]);
}
// Driver Code
let D = 15;
let M = "January";
// Function Call
getDate(D, M);
// This code is contributed by susmitakundugoaldanga
</script>
Producción:
16 July
Tiempo Complejidad: O(1)
Espacio Auxiliar: O(1)
Publicación traducida automáticamente
Artículo escrito por poulami21ghosh y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA