如何使用高斯算法
计算给定年份的复活节日期
给定一个整数 Y ，表示一年，作为输入，任务是找到该年的复活节日期。
示例:
输入: Y = 2020 输出: 2020-04-12 说明: 2020 年复活节是 4 月 12 日 因此，复活节日期将是 2020-04-12
输入: Y = 1991 输出: 1991-03-30 解释: 1991 年，复活节是 3 月 30 日 因此，复活节日期将是 1991-03-30
方法: 高斯复活节算法用于轻松计算一年的复活节日期。卡尔·弗里德里希·高斯首先想到了这个算法。关于高斯是如何得出这个算法的，没有适当的解释，但是这个算法的实现被证明是非常精确的。 高斯复活节算法的详细解释如下:
- 首先，计算年 Y 在美托周期中的位置。
A = Y mod 19
- 现在，根据儒略历找出闰日的数字。
B = Y mod 4
- 然后，让我们考虑一下，非闰年比 52 周长一天。
C = Y mod 7
- M 取决于一年中的世纪 Y 。对于 19 世纪，M = 23。对于 21 世纪，M = 24 等等。 使用以下关系计算:
P =楼板(y/100) q =((13+8 * p)/25) m =(15–q+p-(p/4))mod 30
- 儒略历和公历的闰日数之差由下式给出:
N = （4 + P – （P / 4）） mod 7
- 为了找到逾越节满月的日期，3 月 21 日的天数由下式给出:
D = (19*A + M) mod 30
- 并且，从逾越节满月到下一个星期日的天数由下式给出:
E = (N + 2B + 4C + 6*D) mod 7
- 因此，使用 D 和 E ，复活节周日的日期将是3 月(22 + D + E) 。如果这个数字大于 31，那么我们就移到 4 月。
- 现在农历月份不是正好 30 天而是少了一点 30 天。因此，为了消除这种不一致，遵循以下情况:
if (D == 29)和(E == 6) 返回“4 月 19 日” 否则 if (D == 28)和(E == 6) 返回“4 月 18 日”
下面是上述方法的实现:
C++
// C++ program for the
// above approach
#include <iostream>
#include <math.h>
using namespace std;
// Function calculates and prints
// easter date for given year Y
void gaussEaster(int Y)
{
float A, B, C, P, Q,
M, N, D, E;
// All calculations done
// on the basis of
// Gauss Easter Algorithm
A = Y % 19;
B = Y % 4;
C = Y % 7;
P = (float)floor(Y / 100);
Q = (float)floor((13 + 8 * P) / 25);
M = (int)(15 - Q + P - P / 4) % 30;
N = (int)(4 + P - P / 4) % 7;
D = (int)(19 * A + M) % 30;
E = (int)(2 * B + 4 * C + 6 * D + N) % 7;
int days = (int)(22 + D + E);
// A corner case,
// when D is 29
if ((D == 29) && (E == 6)) {
cout << Y << "-04-19";
return;
}
// Another corner case,
// when D is 28
else if ((D == 28) && (E == 6)) {
cout << Y << "-04-18";
return;
}
else {
// If days > 31, move to April
// April = 4th Month
if (days > 31) {
cout << Y << "-04-"
<< (days - 31);
return;
}
else {
// Otherwise, stay on March
// March = 3rd Month
cout << Y << "-03-"
<< days;
return;
}
}
}
// Driver Code
int main()
{
int Y = 2020;
gaussEaster(Y);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program for the
// above approach
import java.util.Date;
import java.util.Scanner;
// Function calculates and prints
// easter date for given year Y
public class GaussEaster {
static void gaussEaster(int Y)
{
float A, B, C, P, Q,
M, N, D, E;
// All calculations done
// on the basis of
// Gauss Easter Algorithm
A = Y % 19;
B = Y % 4;
C = Y % 7;
P = (float)Math.floor(Y / 100);
Q = (float)Math.floor(
(13 + 8 * P) / 25);
M = (15 - Q + P - P / 4) % 30;
N = (4 + P - P / 4) % 7;
D = (19 * A + M) % 30;
E = (2 * B + 4 * C + 6 * D + N) % 7;
int days = (int)(22 + D + E);
// A corner case,
// when D is 29
if ((D == 29) && (E == 6)) {
System.out.println(Y + "-04"
+ "-19");
return;
}
// Another corner case,
// when D is 28
else if ((D == 28) && (E == 6)) {
System.out.println(Y + "-04"
+ "-18");
return;
}
else {
// If days > 31, move to April
// April = 4th Month
if (days > 31) {
System.out.println(Y + "-04-"
+ (days - 31));
return;
}
// Otherwise, stay on March
// March = 3rd Month
else {
System.out.println(Y + "-03-"
+ days);
return;
}
}
}
// Driver code
public static void main(String[] args)
{
int Y = 2020;
gaussEaster(Y);
}
}
Python 3
# Python3 program for the
# above approach
import math
# Function calculates and prints
# easter date for given year Y
def gaussEaster(Y):
# All calculations done
# on the basis of
# Gauss Easter Algorithm
A = Y % 19
B = Y % 4
C = Y % 7
P = math.floor(Y / 100)
Q = math.floor((13 + 8 * P) / 25)
M = (15 - Q + P - P // 4) % 30
N = (4 + P - P // 4) % 7
D = (19 * A + M) % 30
E = (2 * B + 4 * C + 6 * D + N) % 7
days = (22 + D + E)
# A corner case,
# when D is 29
if ((D == 29) and (E == 6)):
print(Y, "-04-19")
return
# Another corner case,
# when D is 28
elif ((D == 28) and (E == 6)):
print(Y, "-04-18")
return
else:
# If days > 31, move to April
# April = 4th Month
if (days > 31):
print(Y, "-04-", (days - 31))
return
else:
# Otherwise, stay on March
# March = 3rd Month
print(Y, "-03-", days)
return
# Driver Code
Y = 2020
gaussEaster(Y)
# This code is contributed by code_hunt
C
// C# program for the
// above approach
using System;
class GFG{
// Function calculates and prints
// easter date for given year Y
static void gaussEaster(int Y)
{
float A, B, C, P, Q,
M, N, D, E;
// All calculations done
// on the basis of
// Gauss Easter Algorithm
A = Y % 19;
B = Y % 4;
C = Y % 7;
P = (float)(Y / 100);
Q = (float)(
(13 + 8 * P) / 25);
M = (15 - Q + P - P / 4) % 30;
N = (4 + P - P / 4) % 7;
D = (19 * A + M) % 30;
E = (2 * B + 4 * C + 6 * D + N) % 7;
int days = (int)(22 + D + E);
// A corner case,
// when D is 29
if ((D == 29) && (E == 6))
{
Console.Write(Y + "-04" + "-19");
return;
}
// Another corner case,
// when D is 28
else if ((D == 28) && (E == 6))
{
Console.Write(Y + "-04" + "-18");
return;
}
else
{
// If days > 31, move to April
// April = 4th Month
if (days > 31)
{
Console.Write(Y + "-04-" +
(days - 31));
return;
}
// Otherwise, stay on March
// March = 3rd Month
else
{
Console.Write(Y + "-03-" + days);
return;
}
}
}
// Driver code
public static void Main(string[] args)
{
int Y = 2020;
gaussEaster(Y);
}
}
// This code is contributed by Ritik Bansal
java 描述语言
<script>
// Javascript implementation for the above approach
// Function calculates and prlets
// easter date for given year Y
function gaussEaster(Y)
{
let A, B, C, P, Q,
M, N, D, E;
// All calculations done
// on the basis of
// Gauss Easter Algorithm
A = Y % 19;
B = Y % 4;
C = Y % 7;
P = Math.floor(Y / 100);
Q = Math.floor(
(13 + 8 * P) / 25);
M = (15 - Q + P - P / 4) % 30;
N = (4 + P - P / 4) % 7;
D = (19 * A + M) % 30;
E = (2 * B + 4 * C + 6 * D + N) % 7;
let days = (22 + D + E);
// A corner case,
// when D is 29
if ((D == 29) && (E == 6)) {
document.write(Y + "-04"
+ "-19");
return;
}
// Another corner case,
// when D is 28
else if ((D == 28) && (E == 6)) {
document.write(Y + "-04"
+ "-18");
return;
}
else {
// If days > 31, move to April
// April = 4th Month
if (days > 31) {
document.write(Y + "-04-"
+ (days - 31));
return;
}
// Otherwise, stay on March
// March = 3rd Month
else {
document.write(Y + "-03-"
+ days);
return;
}
}
}
// Driver Code
let Y = 2020;
gaussEaster(Y);
</script>
Output:
2020-04-12
时间复杂度: O(1) 辅助空间: O(1)
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