查找下一个相同的日历年
给你一个 Y 年，找到下一个与 Y 相同的日历年。 例:
Input : 2017
Output : 2023
Input : 2018
Output : 2029
如果满足以下两个条件，一年 x 与给定的上一年 y 相同。
- x 和 y 从同一天开始。
- 如果 y 是闰年，那么 x 也是。如果 y 不是闰年，那么 x 也不是。
想法是逐一检查所有年份(从明年开始)。我们记录前进的天数。如果总移动天数为 7 天，则当年从同一天开始。我们还检查当年的闰是否与 y 相同。如果两个条件都满足，我们就返回当年。
C++
// C++ program to find next identical year
#include<iostream>
using namespace std;
// Function for finding extra days of year
// more than complete weeks
int extraDays(int y)
{
// If current year is a leap year, then
// it number of weekdays move ahead by
// 2 in terms of weekdays.
if (y%400==0 || y%100!=0 && y%4==0)
return 2;
// Else number of weekdays move ahead
// by 1.
return 1;
}
// Returns next identical year.
int nextYear(int y)
{
// Find number of days moved ahead by y
int days = extraDays(y);
// Start from next year
int x = y + 1;
// Count total number of weekdays
// moved ahead so far.
for (int sum=0; ; x++)
{
sum = (sum + extraDays(x)) % 7;
// If sum is divisible by 7 and leap-ness
// of x is same as y, return x.
if ( sum==0 && (extraDays(x) == days))
return x;
}
return x;
}
// driver program
int main()
{
int y = 2018;
cout << nextYear(y);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program to find next identical year
class GFG {
// Function for finding extra days of year
// more than complete weeks
static int extraDays(int y)
{
// If current year is a leap year, then
// it number of weekdays move ahead by
// 2 in terms of weekdays.
if (y % 400 == 0 || y % 100 != 0 && y % 4 == 0)
return 2;
// Else number of weekdays move ahead
// by 1.
return 1;
}
// Returns next identical year.
static int nextYear(int y)
{
// Find number of days moved ahead by y
int days = extraDays(y);
// Start from next year
int x = y + 1;
// Count total number of weekdays
// moved ahead so far.
for (int sum = 0; ; x++)
{
sum = (sum + extraDays(x)) % 7;
// If sum is divisible by 7 and leap-ness
// of x is same as y, return x.
if ( sum == 0 && (extraDays(x) == days))
return x;
}
}
// Driver code
public static void main(String[] args)
{
int y = 2018;
System.out.println(nextYear(y));
}
}
/* This code contributed by PrinciRaj1992 */
Python 3
# Python3 program to find next identical year
# Function for finding extra days of year
# more than complete weeks
def extraDays(y) :
# If current year is a leap year, then
# it number of weekdays move ahead by
# 2 in terms of weekdays.
if (y % 400 == 0 or y % 100 != 0 and y % 4 == 0) :
return 2
# Else number of weekdays move ahead
# by 1.
return 1
# Returns next identical year.
def nextYear(y) :
# Find number of days moved ahead by y
days = extraDays(y)
# Start from next year
x = y + 1
# Count total number of weekdays
# moved ahead so far.
Sum = 0
while(True) :
Sum = (Sum + extraDays(x)) % 7
# If sum is divisible by 7 and leap-ness
# of x is same as y, return x.
if ( Sum == 0 and (extraDays(x) == days)) :
return x
x += 1
return x
y = 2018
print(nextYear(y))
# This code is contributed by mukesh07.
C
// C# program to find next identical year
using System;
class GFG
{
// Function for finding extra days of year
// more than complete weeks
static int extraDays(int y)
{
// If current year is a leap year, then
// it number of weekdays move ahead by
// 2 in terms of weekdays.
if (y % 400 == 0 || y % 100 != 0 && y % 4 == 0)
return 2;
// Else number of weekdays move ahead
// by 1.
return 1;
}
// Returns next identical year.
static int nextYear(int y)
{
// Find number of days moved ahead by y
int days = extraDays(y);
// Start from next year
int x = y + 1;
// Count total number of weekdays
// moved ahead so far.
for (int sum = 0; ; x++)
{
sum = (sum + extraDays(x)) % 7;
// If sum is divisible by 7 and leap-ness
// of x is same as y, return x.
if ( sum == 0 && (extraDays(x) == days))
return x;
}
}
// Driver code
public static void Main(String[] args)
{
int y = 2018;
Console.WriteLine(nextYear(y));
}
}
// This code has been contributed by 29AjayKumar
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP program to find
// next identical year
// Function for finding extra days
// of year more than complete weeks
function extraDays($y)
{
// If current year is a leap year,
// then number of weekdays move
// ahead by 2 in terms of weekdays.
if ($y % 400 == 0 ||
$y % 100 != 0 &&
$y % 4 == 0)
return 2;
// Else number of weekdays
// move ahead by 1.
return 1;
}
// Returns next identical year.
function nextYear($y)
{
// Find number of days
// moved ahead by y
$days = extraDays($y);
// Start from next year
$x = $y + 1;
// Count total number of weekdays
// moved ahead so far.
for ($sum = 0; ; $x++)
{
$sum = ($sum + extraDays($x)) % 7;
// If sum is divisible by 7
// and leap-ness of x is
// same as y, return x.
if ( $sum == 0 && (extraDays($x) == $days))
return $x;
}
return $x;
}
// Driver Code
$y = 2018;
echo nextYear($y);
// This code is contributed by aj_36
?>
java 描述语言
<script>
// JavaScript program for the above approach
// Function for finding extra days of year
// more than complete weeks
function extraDays(y)
{
// If current year is a leap year, then
// it number of weekdays move ahead by
// 2 in terms of weekdays.
if (y % 400 == 0 || y % 100 != 0 && y % 4 == 0)
return 2;
// Else number of weekdays move ahead
// by 1.
return 1;
}
// Returns next identical year.
function nextYear(y)
{
// Find number of days moved ahead by y
let days = extraDays(y);
// Start from next year
let x = y + 1;
// Count total number of weekdays
// moved ahead so far.
for (let sum = 0; ; x++)
{
sum = (sum + extraDays(x)) % 7;
// If sum is divisible by 7 and leap-ness
// of x is same as y, return x.
if ( sum == 0 && (extraDays(x) == days))
return x;
}
}
// Driver Code
let y = 2018;
document.write(nextYear(y));
// This code is contributed by susmitakundugoaldanga.
</script>
输出:
2029
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