从给定日期查找下半年之后的日期

给定一个正整数D和一个字符串M代表闰年的日期和月份，任务是找到下半年之后的日期。

例子：

Input: D = 15, M = “January”
Output: 16 July
Explanation: The date from the 15th of January to the next half year is 16th of July.

Input: D = 10, M = “October”
Output: 10 April

方法：由于闰年包含366 天，因此可以通过将当前日期增加183 天后查找数据来解决给定的问题。请按照以下步骤解决问题：

• 将每个月的天数存储在该数组中，例如days[] 。
• 初始化一个变量，比如curMonth为M ，以存储当月的索引。
• 初始化一个变量，比如curDate为D ，以存储当前日期。
• 初始化一个变量，比如count为183 ，表示要增加的天数。
• 迭代直到count 的值为正，然后执行以下步骤：
  • 如果计数值为正，CURDATE是在curMonth大部分天数则减1计数的值和递增1 CURDATE的价值。
  • 如果count 的值为0，则跳出循环。
  • 按(curMonth + 1)%12更新curMonth的值。
• 完成上述步骤后，打印出curDate和curMonth对应的日期作为结果。

下面是上述方法的实现：

C++

// C++ program for the above approach
 
#include 
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

 
// 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
 

输出：

16 July

时间复杂度： O(1)
辅助空间： O(1)