Dada una fecha y un entero positivo x . La tarea es encontrar la fecha después de agregar x días a la fecha dada
Ejemplos:
Input : d1 = 14, m1 = 5, y1 = 2017
Days to be added x = 10
Output : d2 = 24, m2 = 5, y2 = 2017
Input : d1 = 14, m1 = 3, y1 = 2015
Days to be added x = 466
Output : d2 = 14, m2 = 3, y2 = 2016
1) Sea la fecha dada d1, m1 e y1. Encuentre la compensación (número de días transcurridos desde el comienzo hasta la fecha dada) de un año determinado (consulte offsetDays() a continuación)
2) Deje que la compensación encontrada en el paso anterior sea offset1. Encuentre el año de resultado y2 y la compensación del año de resultado compensación2 (Consulte el código resaltado a continuación)
3) Encuentre los días y meses de la compensación2 y y2. (Consulte revoffsetDays() a continuación).
A continuación se muestra la implementación de los pasos anteriores.
C++
// C++ program to find date after adding
// given number of days.
#include<bits/stdc++.h>
using namespace std;
// Return if year is leap year or not.
bool isLeap(int y)
{
if (y%100 != 0 && y%4 == 0 || y %400 == 0)
return true;
return false;
}
// Given a date, returns number of days elapsed
// from the beginning of the current year (1st
// jan).
int offsetDays(int d, int m, int y)
{
int offset = d;
switch (m - 1)
{
case 11:
offset += 30;
case 10:
offset += 31;
case 9:
offset += 30;
case 8:
offset += 31;
case 7:
offset += 31;
case 6:
offset += 30;
case 5:
offset += 31;
case 4:
offset += 30;
case 3:
offset += 31;
case 2:
offset += 28;
case 1:
offset += 31;
}
if (isLeap(y) && m > 2)
offset += 1;
return offset;
}
// Given a year and days elapsed in it, finds
// date by storing results in d and m.
void revoffsetDays(int offset, int y, int *d, int *m)
{
int month[13] = { 0, 31, 28, 31, 30, 31, 30,
31, 31, 30, 31, 30, 31 };
if (isLeap(y))
month[2] = 29;
int i;
for (i = 1; i <= 12; i++)
{
if (offset <= month[i])
break;
offset = offset - month[i];
}
*d = offset;
*m = i;
}
// Add x days to the given date.
void addDays(int d1, int m1, int y1, int x)
{
int offset1 = offsetDays(d1, m1, y1);
int remDays = isLeap(y1)?(366-offset1):(365-offset1);
// y2 is going to store result year and
// offset2 is going to store offset days
// in result year.
int y2, offset2;
if (x <= remDays)
{
y2 = y1;
offset2 = offset1 + x;
}
else
{
// x may store thousands of days.
// We find correct year and offset
// in the year.
x -= remDays;
y2 = y1 + 1;
int y2days = isLeap(y2)?366:365;
while (x >= y2days)
{
x -= y2days;
y2++;
y2days = isLeap(y2)?366:365;
}
offset2 = x;
}
// Find values of day and month from
// offset of result year.
int m2, d2;
revoffsetDays(offset2, y2, &d2, &m2);
cout << "d2 = " << d2 << ", m2 = " << m2
<< ", y2 = " << y2;
}
// Driven Program
int main()
{
int d = 14, m = 3, y = 2015;
int x = 366;
addDays(d, m, y, x);
return 0;
}
Java
// Java program to find date after adding
// given number of days.
class GFG
{
// Find values of day and month from
// offset of result year.
static int m2, d2;
// Return if year is leap year or not.
static boolean isLeap(int y)
{
if (y % 100 != 0 && y % 4 == 0 || y % 400 == 0)
return true;
return false;
}
// Given a date, returns number of days elapsed
// from the beginning of the current year (1st
// jan).
static int offsetDays(int d, int m, int y)
{
int offset = d;
if(m - 1 == 11)
offset += 335;
if(m - 1 == 10)
offset += 304;
if(m - 1 == 9)
offset += 273;
if(m - 1 == 8)
offset += 243;
if(m - 1 == 7)
offset += 212;
if(m - 1 == 6)
offset += 181;
if(m - 1 == 5)
offset += 151;
if(m - 1 == 4)
offset += 120;
if(m - 1 == 3)
offset += 90;
if(m - 1 == 2)
offset += 59;
if(m - 1 == 1)
offset += 31;
if (isLeap(y) && m > 2)
offset += 1;
return offset;
}
// Given a year and days elapsed in it, finds
// date by storing results in d and m.
static void revoffsetDays(int offset, int y)
{
int []month={ 0, 31, 28, 31, 30, 31, 30,
31, 31, 30, 31, 30, 31 };
if (isLeap(y))
month[2] = 29;
int i;
for (i = 1; i <= 12; i++)
{
if (offset <= month[i])
break;
offset = offset - month[i];
}
d2 = offset;
m2 = i;
}
// Add x days to the given date.
static void addDays(int d1, int m1, int y1, int x)
{
int offset1 = offsetDays(d1, m1, y1);
int remDays = isLeap(y1) ? (366 - offset1) : (365 - offset1);
// y2 is going to store result year and
// offset2 is going to store offset days
// in result year.
int y2, offset2 = 0;
if (x <= remDays)
{
y2 = y1;
offset2 =offset1 + x;
}
else
{
// x may store thousands of days.
// We find correct year and offset
// in the year.
x -= remDays;
y2 = y1 + 1;
int y2days = isLeap(y2) ? 366 : 365;
while (x >= y2days)
{
x -= y2days;
y2++;
y2days = isLeap(y2) ? 366 : 365;
}
offset2 = x;
}
revoffsetDays(offset2, y2);
System.out.println("d2 = " + d2 + ", m2 = " +
m2 + ", y2 = " + y2);
}
// Driven Program
public static void main(String[] args)
{
int d = 14, m = 3, y = 2015;
int x = 366;
addDays(d, m, y, x);
}
}
// This code is contributed by mits
Python3
# Python3 program to find date after adding
# given number of days.
# Return if year is leap year or not.
def isLeap(y):
if (y % 100 != 0 and y % 4 == 0 or y % 400 == 0):
return True
return False
# Given a date, returns number of days elapsed
# from the beginning of the current year (1st
# jan).
def offsetDays(d, m, y):
offset = d
switcher = {
10: 30,
9: 31,
8: 30,
7: 31,
6: 31,
5: 30,
4: 31,
3: 30,
2: 31,
1: 28,
0: 31
}
if (isLeap(y) and m > 1):
offset += 1
offset +=switcher.get(m)
return offset
# Given a year and days elapsed in it, finds
# date by storing results in d and m.
def revoffsetDays(offset, y,d, m):
month = [ 0, 31, 28, 31, 30, 31, 30,31, 31, 30, 31, 30, 31 ]
if (isLeap(y)):
month[2] = 29
for i in range(1, 13):
if (offset <= month[i]):
break
offset = offset - month[i]
d[0] = offset
m[0] = i + 1
# Add x days to the given date.
def addDays(d1, m1, y1, x):
offset1 = offsetDays(d1, m1, y1)
if isLeap(y1):
remDays = 366 - offset1
else:
remDays = 365 - offset1
# y2 is going to store result year and
# offset2 is going to store offset days
# in result year.
if (x <= remDays):
y2 = y1
offset2 = offset1 + x
else:
# x may store thousands of days.
# We find correct year and offset
# in the year.
x -= remDays
y2 = y1 + 1
if isLeap(y2):
y2days = 366
else:
y2days = 365
while (x >= y2days):
x -= y2days
y2 += 1
if isLeap(y2):
y2days = 366
else:
y2days = 365
offset2 = x
# Find values of day and month from
# offset of result year.
m2 = [0]
d2 = [0]
revoffsetDays(offset2, y2, d2, m2)
print("d2 = ",*d2,", m2 = ",*m2,", y2 = ",y2,sep="")
# Driven Program
d = 14
m = 3
y = 2015
x = 366
addDays(d, m, y, x)
# This code is contributed by shubhamsingh10
C#
// C# program to find date after adding
// given number of days.
using System;
class GFG{
// Find values of day and month from
// offset of result year.
static int m2, d2;
// Return if year is leap year or not.
static bool isLeap(int y)
{
if (y % 100 != 0 && y % 4 == 0 || y % 400 == 0)
return true;
return false;
}
// Given a date, returns number of days elapsed
// from the beginning of the current year (1st
// jan).
static int offsetDays(int d, int m, int y)
{
int offset = d;
if(m - 1 == 11)
offset += 335;
if(m - 1 == 10)
offset += 304;
if(m - 1 == 9)
offset += 273;
if(m - 1 == 8)
offset += 243;
if(m - 1 == 7)
offset += 212;
if(m - 1 == 6)
offset += 181;
if(m - 1 == 5)
offset += 151;
if(m - 1 == 4)
offset += 120;
if(m - 1 == 3)
offset += 90;
if(m - 1 == 2)
offset += 59;
if(m - 1 == 1)
offset += 31;
if (isLeap(y) && m > 2)
offset += 1;
return offset;
}
// Given a year and days elapsed in it, finds
// date by storing results in d and m.
static void revoffsetDays(int offset, int y)
{
int []month = { 0, 31, 28, 31, 30, 31, 30,
31, 31, 30, 31, 30, 31 };
if (isLeap(y))
month[2] = 29;
int i;
for (i = 1; i <= 12; i++)
{
if (offset <= month[i])
break;
offset = offset - month[i];
}
d2 = offset;
m2 = i;
}
// Add x days to the given date.
static void addDays(int d1, int m1, int y1, int x)
{
int offset1 = offsetDays(d1, m1, y1);
int remDays = isLeap(y1) ? (366 - offset1) : (365 - offset1);
// y2 is going to store result year and
// offset2 is going to store offset days
// in result year.
int y2, offset2 = 0;
if (x <= remDays)
{
y2 = y1;
offset2 = offset1+x;
}
else
{
// x may store thousands of days.
// We find correct year and offset
// in the year.
x -= remDays;
y2 = y1 + 1;
int y2days = isLeap(y2) ? 366 : 365;
while (x >= y2days)
{
x -= y2days;
y2++;
y2days = isLeap(y2)?366:365;
}
offset2 = x;
}
revoffsetDays(offset2, y2);
Console.WriteLine("d2 = " + d2 + ", m2 = " +
m2 + ", y2 = " + y2);
}
// Driven Program
static void Main()
{
int d = 14, m = 3, y = 2015;
int x = 366;
addDays(d, m, y, x);
}
}
// This code is contributed by mits
Javascript
<script>
// Javascript program to find date after
// adding given number of days.
// Find values of day and month from
// offset of result year.
var m2, d2;
// Return if year is leap year or not.
function isLeap(y)
{
if (y % 100 != 0 &&
y % 4 == 0 ||
y % 400 == 0)
return true;
return false;
}
// Given a date, returns number of
// days elapsed from the beginning
// of the current year (1st jan).
function offsetDays(d , m , y)
{
var offset = d;
if (m - 1 == 11)
offset += 335;
if (m - 1 == 10)
offset += 304;
if (m - 1 == 9)
offset += 273;
if (m - 1 == 8)
offset += 243;
if (m - 1 == 7)
offset += 212;
if (m - 1 == 6)
offset += 181;
if (m - 1 == 5)
offset += 151;
if (m - 1 == 4)
offset += 120;
if (m - 1 == 3)
offset += 90;
if (m - 1 == 2)
offset += 59;
if (m - 1 == 1)
offset += 31;
if (isLeap(y) && m > 2)
offset += 1;
return offset;
}
// Given a year and days elapsed in it, finds
// date by storing results in d and m.
function revoffsetDays(offset, y)
{
var month = [ 0, 31, 28, 31, 30, 31, 30,
31, 31, 30, 31, 30, 31 ];
if (isLeap(y))
month[2] = 29;
var i;
for(i = 1; i <= 12; i++)
{
if (offset <= month[i])
break;
offset = offset - month[i];
}
d2 = offset;
m2 = i;
}
// Add x days to the given date.
function addDays(d1 , m1 , y1 , x)
{
var offset1 = offsetDays(d1, m1, y1);
var remDays = isLeap(y1) ? (366 - offset1) :
(365 - offset1);
// y2 is going to store result year and
// offset2 is going to store offset days
// in result year.
var y2, offset2 = 0;
if (x <= remDays)
{
y2 = y1;
offset2 =offset1 + x;
}
else
{
// x may store thousands of days.
// We find correct year and offset
// in the year.
x -= remDays;
y2 = y1 + 1;
var y2days = isLeap(y2) ? 366 : 365;
while (x >= y2days)
{
x -= y2days;
y2++;
y2days = isLeap(y2) ? 366 : 365;
}
offset2 = x;
}
revoffsetDays(offset2, y2);
document.write("d2 = " + d2 +
", m2 = " + m2 +
", y2 = " + y2);
}
// Driver code
var d = 14, m = 3, y = 2015;
var x = 366;
addDays(d, m, y, x);
// This code is contributed by Amit Katiyar
</script>
Producción:
d2 = 14, m2 = 3, y2 = 2016
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Artículo escrito por GeeksforGeeks-1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA