你指的是阴历吧？阴历这样算，需要一个字典表：
var lunarLeapMonthOfYear=function(year){var table=[
0x04bd8,0x04ae0,0x0a570,0x054d5,0x0d260,0x0d950,0x16554,0x056a0,0x09ad0,
0x055d2,0x04ae0,0x0a5b6,0x0a4d0,0x0d250,0x1d255,0x0b540,0x0d6a0,0x0ada2,
0x095b0,0x14977,0x04970,0x0a4b0,0x0b4b5,0x06a50,0x06d40,0x1ab54,0x02b60,
0x09570,0x052f2,0x04970,0x06566,0x0d4a0,0x0ea50,0x06e95,0x05ad0,0x02b60,
0x186e3,0x092e0,0x1c8d7,0x0c950,0x0d4a0,0x1d8a6,0x0b550,0x056a0,0x1a5b4,
0x025d0,0x092d0,0x0d2b2,0x0a950,0x0b557,0x06ca0,0x0b550,0x15355,0x04da0,
0x0a5d0,0x14573,0x052d0,0x0a9a8,0x0e950,0x06aa0,0x0aea6,0x0ab50,0x04b60,
0x0aae4,0x0a570,0x05260,0x0f263,0x0d950,0x05b57,0x056a0,0x096d0,0x04dd5,
0x04ad0,0x0a4d0,0x0d4d4,0x0d250,0x0d558,0x0b540,0x0b5a0,0x195a6,0x095b0,
0x049b0,0x0a974,0x0a4b0,0x0b27a,0x06a50,0x06d40,0x0af46,0x0ab60,0x09570,
0x04af5,0x04970,0x064b0,0x074a3,0x0ea50,0x06b58,0x055c0,0x0ab60,0x096d5,
0x092e0,0x0c960,0x0d954,0x0d4a0,0x0da50,0x07552,0x056a0,0x0abb7,0x025d0,
0x092d0,0x0cab5,0x0a950,0x0b4a0,0x0baa4,0x0ad50,0x055d9,0x04ba0,0x0a5b0,
0x15176,0x052b0,0x0a930,0x07954,0x06aa0,0x0ad50,0x05b52,0x04b60,0x0a6e6,
0x0a4e0,0x0d260,0x0ea65,0x0d530,0x05aa0,0x076a3,0x096d0,0x04bd7,0x04ad0,
0x0a4d0,0x1d0b6,0x0d250,0x0d520,0x0dd45,0x0b5a0,0x056d0,0x055b2,0x049b0,
0x0a577,0x0a4b0,0x0aa50,0x1b255,0x06d20,0x0ada0
]
return table[year-1900]&0xf
}
lunarLeapMonthOfYear(2004)//2004年阴历闰二月
lunarLeapMonthOfYear(2000)//2000年阴历闰四月
lunarLeapMonthOfYear(2001)//2001年没有闰月
其实我也不懂这个算法，网上粘贴的，我觉得可以不去深究这个公式，反正有现成代码：
设：公元年数－1977（或1901）＝4Q＋R
则：阴历日期=14Q+10.6(R+1)+年内日期序数-29.5n
（注:式中Q、R、n均为自然数，R<4）
例：1994年5月7日的阴历日期为：
1994－1977＝17＝4×4＋1
故：Q＝4，R＝1 则：5月7日的阴历日期为：
14×4+10.6(1+1)+(31+28+31+30+7)-29.5n
=204.2- 29.5n
然后用29.5去除204.2得商数6......27.2，6即是n值，余数27即是阴历二十七日。
