从给定日期
找到下半年之后的日期
给定一个正整数 D 和一个代表闰年的日和月的字符串 M ，任务是找到下一个半年后的日期。
示例:
输入: D = 15，M =【一月】 输出:7 月 16 日 说明:1 月 15 日至下半年的日期为 7 月 16 日。
输入: D = 10，M =【十月】 T5输出:**4 月 10 日
方法:由于闰年包含 366 天，给定的问题可以通过将当前日期增加 183 天后找到数据来解决。按照以下步骤解决问题:
- 存储该数组中每个月的天数，比如天【】。
- 初始化一个变量，说 curMonth 为 M ，存储当月的指数。
- 初始化一个变量，说将压缩为 D ，以存储当前日期。
- 初始化一个变量，说计数为 183 ，代表要递增的天数。
- 迭代直到计数的值为正，并执行以下步骤:
- 如果计数的值为正，凝结最多为月的天数，则计数的值递减 1 ，将凝结的值递增 1 。
- 如果计数的值为 0 ，则退出循环。
- 通过 (curMonth + 1)%12 更新 curMonth 的值。
- 完成以上步骤后，打印对应的日期，打印结果为和 curMonth 。
下面是上述方法的实现:
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 语言(一种计算机语言，尤用于创建网站)
// 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
Python 3
# 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
java 描述语言
<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>
Output:
16 July
时间复杂度:O(1) T5辅助空间:** O(1)
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