今天看到一个很有趣的题目，题目描述如下：
根据下列信息计算在1901年1月1日至2000年12月31日间共有多少个星期天落在每月的第一天上？
a) 1900.1.1是星期一
b) 1月，3月，5月，7月，8月，10月和12月是31天
c) 4月，6月，9月和11月是30天
d) 2月是28天，在闰年是29天
e) 公元年数能被4整除且又不能被100整除是闰年
f) 能直接被400整除也是闰年
以下是C语言实现版本：
#include <stdio.h>
#include <stdbool.h>
bool isLeapYear(int year);
// start is the weekday of 1st, January
// return the num of the first day of each month
// is Sunday.
// the start will change into the next year
int getYearNum(int* start, int year);
// Num of days of each month
int days[13] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
int leapdays[13] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
int main(void)
{
int sum = 0;
int start = 1901;
int end = 2000;
int startWeek = 1;
int startYear = 1900;
int i;
for (i = 1; i < 13; ++i)
days[i] += days[i - 1];
for (i = 1; i < 13; ++i)
leapdays[i] += leapdays[i - 1];
for (i = startYear; i < start; ++i)
getYearNum(&startWeek, i);
for (i = start; i <= end; ++i)
sum += getYearNum(&startWeek, i);
printf("%d\n", sum);
return 0;
}
bool isLeapYear(int year)
{
if (year % 4 == 0 && year % 100 != 0)
return true;
else if (year % 400 == 0)
return true;
return false;
}
int getYearNum(int* start, int year)
{
int i;
int count = 0;
int yeardays;
if (isLeapYear(year))
{
yeardays = 366;
for (i = 0; i < 12; ++i)
if ((leapdays[i] % 7 + *start)%7 == 0)
++count;
} else
{
yeardays = 365;
for (i = 0; i < 12; ++i)
if ((days[i] % 7 + *start)%7 == 0)
++count;
}
*start = (yeardays % 7 + *start)%7;
return count;
}
下来是最简单的python:
import calendar sum = 0 startYear = 1901 endYear = 2000 for year in xrange(startYear, endYear + 1): for month in xrange(1, 13): if calendar.monthcalendar(year, month)[0].index(1) == 6: sum = sum + 1 print sum
最近对于JavaScript的网页脚本有点感兴趣，就试着用JavaScript实现了一下，感觉不错，有可视化和跨平台性：
<!DOCTYPE html>
<html lang="zh-cn">
<head>
<meta charset="UTF-8">
<meta name="viewport" content="width=device-width, initial-scale=1">
<title>问题2的解答：Javascript版本</title>
</head>
<body>
<script type="text/javascript">
yearTest = /^[1-9]\d{3}$/;
days = new Array(0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31);
leapdays = new Array(0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31);
for (var i = 1; i < days.length; i++)
days[i] += days[i - 1];
for (var i = 1; i < leapdays.length; i++)
leapdays[i] += leapdays[i - 1];
sum = 0;
function isLeap(year)
{
if (!yearTest.test(year))
return false;
if (year % 4 == 0 && year % 100 != 0)
return true;
if (year % 400 == 0)
return true;
return false;
}
startWeek = 1;
totalDays = 365;
function getResultOfYear(year)
{
var sum = 0;
if (isLeap(year))
{
totalDays = 366;
for (var i = 0; i < leapdays.length - 1; i++)
{
if ((leapdays[i] % 7 + startWeek) % 7 == 0)
{
//alert("year: " + year + " month: " + (i + 1));
sum++;
}
}
startWeek = (totalDays % 7 + startWeek) % 7;
}
else
{
totalDays = 365;
for (var i = 0; i < days.length - 1; i++)
{
if ((days[i] % 7 + startWeek) % 7 == 0)
sum++;
}
startWeek = (totalDays % 7 + startWeek) % 7;
}
return sum;
}
function main()
{
var startYear=document.getElementById("startText").value.replace(/(^\s+)|(\s+$)/g,'');
if (startYear == "" || !yearTest.test(startYear))
{
alert("不合法的起始年份！");
return;
}
if (startYear < 1900)
{
alert("起始年份必须大于等于1900！");
return;
}
var endYear = document.getElementById("endText").value.replace(/(^\s+)|(\s+$)/g,'');
if (endYear == "" || !yearTest.test(endYear))
{
alert("不合法的终止年份！");
return;
}
if (endYear < startYear)
{
alert("终止年份必须大于等于起始年份！");
return;
}
sum = 0;
startWeek = 1;
for (i = 1900; i < startYear; i++)
getResultOfYear(i);
for (i = startYear; i <= endYear; i++)
sum += getResultOfYear(i);
document.getElementById("resultText").innerHTML = "从"
+ startYear + "年1月1日到" + endYear +
"年12月23日有<b>" + sum + "</b>个月的1日是周日。";
}
</script>
请输入起始年份：
<input type="text" id="startText" autofocus onkeydown=
"if(event.keyCode==13){endText.focus()}">
<br/>
请输入终止年份：
<input type="text" id="endText" onkeydown=
"if(event.keyCode==13){ok.click()}">
<br/>
<button id="ok" onclick="main()">确定</button>
<br/>
<p id="resultText"></p>
</body>
</html>
JavaScript在手机上的效果如下：
OK，一个问题，多种语言，各有优劣。同时用多种语言是一种很不错的体验。
时间： 2024-10-01 23:04:38