从以星期一为第一天的基准年中找到给定年份的第一天
原文:https://www . geeksforgeeks . org/find-给定年份的第一天-从基年开始-从星期一开始的第一天/
给定代表两年的两个整数 Y 和 B ，任务是找到一周中的哪一天是一年中的 1ST1 月 Y 假设一年中的 1ST1 月 B 是星期一。
示例:
输入: Y = 2020 B = 1900 输出: 周三 说明: 01/01/2020 是周三考虑到 01/01/1900 是周一
输入: Y = 2020 B = 1905 输出: 星期四 解释: 2020 年 1 月 1 日是星期三假设 1905 年 1 月 1 日是星期一
方法:按照以下步骤解决问题:
- 介于基准年(B) 和年(Y) 之间的总年数等于(Y–1)–B。
- 介于之间的闰年总数=总年数/ 4
- 介于= 总年数–闰年之间的非闰年总数。
- 总天数= 总闰年 366 +非闰年 365 + 1 。
- 因此，Y1 月 1 日为总天数% 7 。
下面是上述方法的实现:
C++14
// C++ Implementation of
// the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to find the day of
// 1st January of Y year
void findDay(int Y, int B)
{
int lyear, rest, totaldays, day;
// Count years between
// years Y and B
Y = (Y - 1) - B;
// Count leap years
lyear = Y / 4;
// Non leap years
rest = Y - lyear;
// Total number of days in the years
// lying between the years Y and B
totaldays = (rest * 365)
+ (lyear * 366) + 1;
// Actual day
day = (totaldays % 7);
if (day == 0)
printf("Monday");
else if (day == 1)
printf("Tuesday");
else if (day == 2)
printf("Wednesday");
else if (day == 3)
printf("Thursday");
else if (day == 4)
printf("Friday");
else if (day == 5)
printf("Saturday");
else if (day == 6)
printf("Sunday");
else
printf("INPUT YEAR IS WRONG!");
}
// Driver Code
int main()
{
int Y = 2020, B = 1900;
findDay(Y, B);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program to implement
// the above approach
import java.util.*;
class GFG
{
// Function to find the day of
// 1st January of Y year
static void findDay(int Y, int B)
{
int lyear, rest, totaldays, day;
// Count years between
// years Y and B
Y = (Y - 1) - B;
// Count leap years
lyear = Y / 4;
// Non leap years
rest = Y - lyear;
// Total number of days in the years
// lying between the years Y and B
totaldays = (rest * 365)
+ (lyear * 366) + 1;
// Actual day
day = (totaldays % 7);
if (day == 0)
System.out.println("Monday");
else if (day == 1)
System.out.println("Tuesday");
else if (day == 2)
System.out.println("Wednesday");
else if (day == 3)
System.out.println("Thursday");
else if (day == 4)
System.out.println("Friday");
else if (day == 5)
System.out.println("Saturday");
else if (day == 6)
System.out.println("Sunday");
else
System.out.println("INPUT YEAR IS WRONG!");
}
// Driver code
public static void main(String[] args)
{
int Y = 2020, B = 1900;
findDay(Y, B);
}
}
// This code is contributed by code_hunt.
Python 3
# Python program to implement
# the above approach
# Function to find the day of
# 1st January of Y year
def findDay(Y, B):
lyear, rest, totaldays, day = 0, 0, 0, 0;
# Count years between
# years Y and B
Y = (Y - 1) - B;
# Count leap years
lyear = Y // 4;
# Non leap years
rest = Y - lyear;
# Total number of days in the years
# lying between the years Y and B
totaldays = (rest * 365) + (lyear * 366) + 1;
# Actual day
day = (totaldays % 7);
if (day == 0):
print("Monday");
elif (day == 1):
print("Tuesday");
elif (day == 2):
print("Wednesday");
elif (day == 3):
print("Thursday");
elif (day == 4):
print("Friday");
elif (day == 5):
print("Saturday");
elif (day == 6):
print("Sunday");
else:
print("INPUT YEAR IS WRONG!");
# Driver code
if __name__ == '__main__':
Y = 2020; B = 1900;
findDay(Y, B);
# This code is contributed by 29AjayKumar
C
// C# program to implement
// the above approach
using System;
class GFG
{
// Function to find the day of
// 1st January of Y year
static void findDay(int Y, int B)
{
int lyear, rest, totaldays, day;
// Count years between
// years Y and B
Y = (Y - 1) - B;
// Count leap years
lyear = Y / 4;
// Non leap years
rest = Y - lyear;
// Total number of days in the years
// lying between the years Y and B
totaldays = (rest * 365)
+ (lyear * 366) + 1;
// Actual day
day = (totaldays % 7);
if (day == 0)
Console.WriteLine("Monday");
else if (day == 1)
Console.WriteLine("Tuesday");
else if (day == 2)
Console.WriteLine("Wednesday");
else if (day == 3)
Console.WriteLine("Thursday");
else if (day == 4)
Console.WriteLine("Friday");
else if (day == 5)
Console.WriteLine("Saturday");
else if (day == 6)
Console.WriteLine("Sunday");
else
Console.WriteLine("INPUT YEAR IS WRONG!");
}
// Driver code
static void Main()
{
int Y = 2020, B = 1900;
findDay(Y, B);
}
}
// This code is contribute by susmitakundugoaldanga
java 描述语言
<script>
// Javascript program of the above approach
// Function to find the day of
// 1st January of Y year
function findDay(Y, B)
{
let lyear, rest, totaldays, day;
// Count years between
// years Y and B
Y = (Y - 1) - B;
// Count leap years
lyear = Math.floor(Y / 4);
// Non leap years
rest = Y - lyear;
// Total number of days in the years
// lying between the years Y and B
totaldays = (rest * 365)
+ (lyear * 366) + 1;
// Actual day
day = (totaldays % 7);
if (day == 0)
document.write("Monday");
else if (day == 1)
document.write("Tuesday");
else if (day == 2)
document.write("Wednesday");
else if (day == 3)
document.write("Thursday");
else if (day == 4)
document.write("Friday");
else if (day == 5)
document.write("Saturday");
else if (day == 6)
document.write("Sunday");
else
document.write("INPUT YEAR IS WRONG!");
}
// Driver Code
// Given array
let Y = 2020, B = 1900;
findDay(Y, B);
</script>
Output:
Wednesday
时间复杂度:O(1) T5辅助空间:** O(1)
版权属于：月萌API www.moonapi.com，转载请注明出处
