编写一个程序来计算复活节星期日的日期。复活节星期日是春天第一个满月后的第一个星期天。 使用数学家Carl Friedrich Gauss在1800年发明的算法：
y成为年份（例如1800或2001年）y除以19并调用余下的a。忽略商。y除以100以获得商b和其余c。b除以4以获得商d和其余e。8 * b + 13除以25以获得商g。忽略其余部分。19 * a + b - d - g + 15除以30以获得余额h。忽视
商。c除以4以获得商j和其余k。a + 11 * h除以319以获得商m。忽略其余部分。2 * e + 2 * j - k - h + m + 32除以7以获得余额r。忽略商。h - m + r + 90除以25以获得商n。忽略了
其余部分。h - m + r + n + 19除以32，以获得剩余的p。忽略了
商。然后复活节属于p个月的n天。
例如，如果y是2001：
a = 6
b = 20
c = 1
d = 5
e = 0
g = 6
h = 18
j = 0
k = 1
m = 0
r = 6
n = 4
p = 15
因此，在2001年，复活节星期日在4月15日下降。
确保提示用户一年并让用户输入年份。另外，请确保使用描述值输出的相应消息输出p和n的值。
我把它放到Java代码中有点麻烦。这是我尝试过的：
import java.util.Scanner;
public class Easter {
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
int y = 2014;
int a = y % 19;
int b = y / 100;
int c = y % 100;
int d = b / 4;
int e = b % 4;
int g = (8 * b + 13) / 25;
int h = (19 * a + b - d - g + 15) % 30;
int j = c / 4;
int k = c % 4;
int m = (a + 11 * h) / 319;
int r = (2 * e + 2 * j - k - h + m + 32) % 7;
int n = (h - m + r + 90) / 25;
int p = (h - m + r + n + 19) % 32;
getEasterSundayMonth = n;
System.out.println("Month: " + Easter.getEasterSundayMonth());
}
}
这就是我所拥有的。我不知道如何分配内容，比如我试图让getEasterSundayMonth等于n的值，非常确定它不对。我从哪里开始？
答案 0 :(得分：8)
试试这个：
import java.util.Scanner;
class Easter
{
public static void main(String[] args)
{
System.out.print("Please enter a year to calculate Easter Sunday\n>");
Scanner s = new Scanner(System.in);
int inputted = getResult(s);
while(inputted <= 0)
{
System.out.print("Expected a positive year. Please try again:\n>");
inputted = getResult(s);
}
System.out.println(getEasterSundayDate(inputted));
}
private static int getResult(Scanner s)
{
while(!s.hasNextInt())
{
System.out.print("Expected a valid year. Please try again:\n>");
s.nextLine();
}
return s.nextInt();
}
public static String getEasterSundayDate(int year)
{
int a = year % 19,
b = year / 100,
c = year % 100,
d = b / 4,
e = b % 4,
g = (8 * b + 13) / 25,
h = (19 * a + b - d - g + 15) % 30,
j = c / 4,
k = c % 4,
m = (a + 11 * h) / 319,
r = (2 * e + 2 * j - k - h + m + 32) % 7,
n = (h - m + r + 90) / 25,
p = (h - m + r + n + 19) % 32;
String result;
switch(n)
{
case 1:
result = "January ";
break;
case 2:
result = "February ";
break;
case 3:
result = "March ";
break;
case 4:
result = "April ";
break;
case 5:
result = "May ";
break;
case 6:
result = "June ";
break;
case 7:
result = "July ";
break;
case 8:
result = "August ";
break;
case 9:
result = "September ";
break;
case 10:
result = "October ";
break;
case 11:
result = "November ";
break;
case 12:
result = "December ";
break;
default:
result = "error";
}
return result + p;
}
}
输入2001会导致April 15作为输出。
答案 1 :(得分：4)
你离计划不起作用还远远不够。你还有两件事需要做。
使用Scanner提示用户输入的技巧是创建一个while循环来测试用户输入的每一行，并不断重复直到它看到合法值。
而不是硬编码y = 2014;（或其他），你想做这样的事情：
Scanner input = new Scanner(System.in);
int y = -1; // No easter back in B.C.
while (y < 0) {
System.out.println("Please enter a year (integer greater than zero)");
if (input.hasNextInt()) { // check to see if the user entered a number
y = input.nextInt(); // if so, read it
}
input.nextLine(); // advance the scanner to the next line of input
}
在这种情况下，每次用户输入数字时，y仍为-1并且循环继续。
您已经正确地进行了所有计算，因此要结束您的计划，您只需要输出月/日。
我不打算尝试将计算提取到辅助方法中。只需在main()：
int a = y % 19;
int b = y / 100;
...
int n = (h - m + r + 90) / 25;
int p = (h - m + r + n + 19) % 32;
System.out.println("In the year " + y + " Easter with fall on day " + p + " of month " + n);
答案 2 :(得分：1)
VisualBasic Excel VBA Code
Function Easter_Sunday(Year)
k = Int(Year / 100) 'the secular number
s = 2 - Int((3 * k + 3) / 4) 'the secular sun control
m = 15 + Int((3 * k + 3) / 4) - Int((8 * k + 13) / 25) 'the secular moon circuit
a = Year Mod 19 'the lunar parameter
d = (19 * a + m) Mod 30 'the seed for the first full moon in spring
r = Int(d / 29) + (Int(d / 28) - Int(d / 29)) * Int(a / 11) 'the calendar correction amount
EB = 21 + d - r 'the Easter border
FS = 7 - (Year + Int(Year / 4) + s) Mod 7 'the first Sunday in March
ED = 7 - (EB - FS) Mod 7 'Easter distance in days
ESM = EB + ED - 1 'the date of Easter Sunday as the March date
'--------------------------------------- "For Years (1900-9999)" ------------------------------
Easter_Sunday = DateValue(1 & "." & 3 & "." & Year) + ESM
'--------------------------------------- "or for all Years" -----------------------------------
Month_ = 3 + Int(ESM / 31) 'Month March or April
ES = 1 + ESM Mod 31 'Eastersunday
Easter_Sunday = ("Su, " & Year & "." & Format(Month_, "00") & "." & Format(ES, "00"))
End Function
答案 3 :(得分：0)
万一有人在Typescript中寻找算法的更新版本（NY Anonymous Gregorian）...
easterDate() {
var currentYear = new Date().getFullYear();
var a = Math.floor(currentYear % 19);
var b = Math.floor(currentYear / 100);
var c = Math.floor(currentYear % 100);
var d = Math.floor(b / 4);
var e = Math.floor(b % 4);
var f = Math.floor((b + 8) / 25);
var g = Math.floor((b - f + 1) / 3);
var h = Math.floor((19 * a + b - d - g + 15) % 30);
var i = Math.floor(c / 4);
var k = Math.floor(c % 4);
var l = Math.floor((32 + 2 * e + 2 * i - h - k) % 7);
var m = Math.floor((a + 11 * h + 22 * l) / 451);
var n = Math.floor((h + l - 7 * m + 114) / 31);
var p = Math.floor(((h + l - 7 * m + 114) % 31) + 1);
// console.log('a: ' + a + ' b: ' + b + ' c: ' + c + ' d: ' + d + ' e: ' + e);
// console.log('f: ' + f + ' g: ' + g + ' h: ' + h + ' i: ' + i + ' k: ' + k);
// console.log('l: ' + l + ' m: ' + m + ' n: ' + n + ' p: ' + p);
// console.log("In the year " + currentYear + " Easter with fall on day " + p + " of month " + n);
var month = n.toString();
while (month.length < 2) month = "0" + month;
var day = p.toString();
while (day.length < 2) day = "0" + day;
var dateString = currentYear.toString() + '-' + month + '-' + day + 'T00:00:00';
return new Date(dateString);
}
答案 4 :(得分：0)
...或者在ColdFusion中需要它：
<!---
Function to get the easter date adopted to ColdFusion as of
https://stackoverflow.com/questions/26022233/calculate-the-date-of-easter-sunday
--->
<cffunction name="getEasterDate">
<cfargument name="year" required="true">
<cfscript>
var currentYear = arguments["year"];
var a = floor(currentYear % 19);
var b = floor(currentYear / 100);
var c = floor(currentYear % 100);
var d = floor(b / 4);
var e = floor(b % 4);
var f = floor((b + 8) / 25);
var g = floor((b - f + 1) / 3);
var h = floor((19 * a + b - d - g + 15) % 30);
var i = floor(c / 4);
var k = floor(c % 4);
var l = floor((32 + 2 * e + 2 * i - h - k) % 7);
var m = floor((a + 11 * h + 22 * l) / 451);
var n = floor((h + l - 7 * m + 114) / 31);
var p = floor(((h + l - 7 * m + 114) % 31) + 1);
var month = n;
if (len(month) lt 2) {
month = "0" & month;
}
var day = p;
if (len(day) lt 2) {
day = "0" & day;
}
var dateString = day & '.' & month & '.' & currentYear;
return dateString;
</cfscript>
</cffunction>
答案 5 :(得分：0)
/**
* Orthodox easter formula, invented back in 1800 by famous Carl Friedrich Gauss.
*/
public static LocalDate getEasterSundayDate(int year) {
int a = year % 19,
b = year / 100,
c = year % 100,
d = b / 4,
e = b % 4,
g = (8 * b + 13) / 25,
h = (19 * a + b - d - g + 15) % 30,
j = c / 4,
k = c % 4,
m = (a + 11 * h) / 319,
r = (2 * e + 2 * j - k - h + m + 32) % 7,
month = (h - m + r + 90) / 25,
day = (h - m + r + month + 19) % 32;
return LocalDate.of( year, month, day );
}