【原创】VBA（实验5）VBA写干支纪年简单版: 计算阳历年的阴历年份
什么是天干地支
https://zhidao.baidu.com/question/440519452.html
方法1:1个特直接特别笨的办法：先写序列再&
- 直接把天干地支都写成60个循环。
- 天干*6 复制6份
- 地支*5 复制5份
- 然后，两个数组这样对着咬合起来，
- 剩下的就是把每列的文本组合在一起
- 但这样的办法感觉很挫
| 甲 | 乙 | 丙 | 丁 | 戊 | 己 | 庚 | 辛 | 壬 | 癸 | 甲 | 乙 | 丙 | 丁 | 戊 | 己 | 庚 | 辛 | 壬 | 癸 | 甲 | 乙 | 丙 | 丁 | 戊 | 己 | 庚 | 辛 | 壬 | 癸 | 甲 | 乙 | 丙 | 丁 | 戊 | 己 | 庚 | 辛 | 壬 | 癸 | 甲 | 乙 | 丙 | 丁 | 戊 | 己 | 庚 | 辛 | 壬 | 癸 | 甲 | 乙 | 丙 | 丁 | 戊 | 己 | 庚 | 辛 | 壬 | 癸 |
| 子 | 丑 | 寅 | 卯 | 辰 | 巳 | 午 | 未 | 申 | 酉 | 戌 | 亥 | 子 | 丑 | 寅 | 卯 | 辰 | 巳 | 午 | 未 | 申 | 酉 | 戌 | 亥 | 子 | 丑 | 寅 | 卯 | 辰 | 巳 | 午 | 未 | 申 | 酉 | 戌 | 亥 | 子 | 丑 | 寅 | 卯 | 辰 | 巳 | 午 | 未 | 申 | 酉 | 戌 | 亥 | 子 | 丑 | 寅 | 卯 | 辰 | 巳 | 午 | 未 | 申 | 酉 | 戌 | 亥 |
| 甲子 | 乙丑 | 丙寅 | 丁卯 | 戊辰 | 己巳 | 庚午 | 辛未 | 壬申 | 癸酉 | 甲戌 | 乙亥 | 丙子 | 丁丑 | 戊寅 | 己卯 | 庚辰 | 辛巳 | 壬午 | 癸未 | 甲申 | 乙酉 | 丙戌 | 丁亥 | 戊子 | 己丑 | 庚寅 | 辛卯 | 壬辰 | 癸巳 | 甲午 | 乙未 | 丙申 | 丁酉 | 戊戌 | 己亥 | 庚子 | 辛丑 | 壬寅 | 癸卯 | 甲辰 | 乙巳 | 丙午 | 丁未 | 戊申 | 己酉 | 庚戌 | 辛亥 | 壬子 | 癸丑 | 甲寅 | 乙卯 | 丙辰 | 丁巳 | 戊午 | 己未 | 庚申 | 辛酉 | 壬戌 | 癸亥 |
Sub ganzhi()
For i = 1 To 60
Cells(3, i) = Cells(1, i) + Cells(2, i)
Next i
End Sub
方法2：用了系统函数lcm，然后自动排出新的2个60的数组行
- 这个我自己写的
- 但是还是觉得挺2的
- 这2个新的60的数组非要写出来吗？不能直接存在数组里？ 数组能*2这种倍数吗？python可以的
| 庚 | 辛 | 壬 | 癸 | 甲 | 乙 | 丙 | 丁 | 戊 | 己 | ||||||||||||||||||||||||||||||||||||||||||||||||||
| 申 | 酉 | 戌 | 亥 | 子 | 丑 | 寅 | 卯 | 辰 | 巳 | 午 | 未 | ||||||||||||||||||||||||||||||||||||||||||||||||
| 庚 | 辛 | 壬 | 癸 | 甲 | 乙 | 丙 | 丁 | 戊 | 己 | 庚 | 辛 | 壬 | 癸 | 甲 | 乙 | 丙 | 丁 | 戊 | 己 | 庚 | 辛 | 壬 | 癸 | 甲 | 乙 | 丙 | 丁 | 戊 | 己 | 庚 | 辛 | 壬 | 癸 | 甲 | 乙 | 丙 | 丁 | 戊 | 己 | 庚 | 辛 | 壬 | 癸 | 甲 | 乙 | 丙 | 丁 | 戊 | 己 | 庚 | 辛 | 壬 | 癸 | 甲 | 乙 | 丙 | 丁 | 戊 | 己 |
| 申 | 酉 | 戌 | 亥 | 子 | 丑 | 寅 | 卯 | 辰 | 巳 | 午 | 未 | 申 | 酉 | 戌 | 亥 | 子 | 丑 | 寅 | 卯 | 辰 | 巳 | 午 | 未 | 申 | 酉 | 戌 | 亥 | 子 | 丑 | 寅 | 卯 | 辰 | 巳 | 午 | 未 | 申 | 酉 | 戌 | 亥 | 子 | 丑 | 寅 | 卯 | 辰 | 巳 | 午 | 未 | 申 | 酉 | 戌 | 亥 | 子 | 丑 | 寅 | 卯 | 辰 | 巳 | 午 | 未 |
| 庚申 | 辛酉 | 壬戌 | 癸亥 | 甲子 | 乙丑 | 丙寅 | 丁卯 | 戊辰 | 己巳 | 庚午 | 辛未 | 壬申 | 癸酉 | 甲戌 | 乙亥 | 丙子 | 丁丑 | 戊寅 | 己卯 | 庚辰 | 辛巳 | 壬午 | 癸未 | 甲申 | 乙酉 | 丙戌 | 丁亥 | 戊子 | 己丑 | 庚寅 | 辛卯 | 壬辰 | 癸巳 | 甲午 | 乙未 | 丙申 | 丁酉 | 戊戌 | 己亥 | 庚子 | 辛丑 | 壬寅 | 癸卯 | 甲辰 | 乙巳 | 丙午 | 丁未 | 戊申 | 己酉 | 庚戌 | 辛亥 | 壬子 | 癸丑 | 甲寅 | 乙卯 | 丙辰 | 丁巳 | 戊午 | 己未 |
| 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | 16 | 17 | 18 | 19 | 20 | 21 | 22 | 23 | 24 | 25 | 26 | 27 | 28 | 29 | 30 | 31 | 32 | 33 | 34 | 35 | 36 | 37 | 38 | 39 | 40 | 41 | 42 | 43 | 44 | 45 | 46 | 47 | 48 | 49 | 50 | 51 | 52 | 53 | 54 | 55 | 56 | 57 | 58 | 59 | 60 |
Sub ganzhi()
Dim x
Dim arr_tiangan(10)
Dim arr_dizhi(12)
a = 1 '运算前必须进行定义
b = 1
x = Sheets("ganzhi3").UsedRange
For i = 1 To 10
arr_tiangan(i) = Cells(1, i)
Next
For j = 1 To 12
arr_dizhi(j) = Cells(2, j)
Next
m = Application.Lcm(10, 12) '取最小公倍数
m_tiangan = m / 10
m_dizhi = m / 12
For i = 1 To m_tiangan
For j = 1 To 10
Cells(3, a) = arr_tiangan(j)
a = a + 1
Next j
Next i
For i = 1 To m_dizhi
For j = 1 To 12
Cells(4, b) = arr_dizhi(j)
b = b + 1 '这里如果继续用a a的值会继承下来？！
Next j
Next i
For i = 1 To 60
Cells(5, i) = Cells(3, i) + Cells(4, i)
Next
思路3：
Sub test_gz()
Dim arr3()
ReDim arr3(1 To 60)
Dim arr10(1 To 60)
Dim arr20(1 To 60)
arr1 = Array("甲", "乙", "丙", "丁", "戊", "己", "庚", "辛", "壬", "癸")
arr2 = Array("子", "丑", "寅", "卯", "辰", "巳", "午", "未", "申", "酉", "戌", "亥")
'K = 1
'For Each I In arr1
' For Each J In arr2
' arr3(K) = I & J
' Debug.Print arr3(K)
' K = K + 1
' Next
'Next
m = 1
N = 1
For I = 1 To 60 / (UBound(arr1) - LBound(arr1) + 1)
For J = LBound(arr1) To UBound(arr1)
arr10(m) = arr1(J)
m = m + 1
Next
Next
For I = 1 To 60 / (UBound(arr2) - LBound(arr2) + 1)
For J = LBound(arr2) To UBound(arr2)
arr20(N) = arr2(J)
N = N + 1
Next
Next
For I = LBound(arr3) To UBound(arr3)
arr3(I) = arr10(I) + arr20(I)
Debug.Print arr3(I)
Next
End Sub
仿造别人的写的也不对
Sub test_ganzhi()
arr1 = Array("甲", "乙", "丙", "丁", "戊", "己", "庚", "辛", "壬", "癸")
arr2 = Array("子", "丑", "寅", "卯", "辰", "巳", "午", "未", "申", "酉", "戌", "亥")
a = ""
K = 0
Do While Not IsNumeric(a)
a = InputBox("请输入一个纯数字的年份")
K = K + 1
If K > 3 Then
MsgBox "你成心输错的吧！重新运行吧"
End If
Loop
'If Int(a) - 1984 >= 0 Then
' b = Int(a) - 1984
'Else
' c = Abs(Int(a) - 1984) Mod 60
' b = 60 - c
'End If
Debug.Print arr1(((Int(a) - 4) Mod 60) Mod 10) & arr2(((Int(a) - 4) Mod 60) Mod 12)
End Sub
参考，下面的是找到的其他人的写法
http://club.excelhome.net/thread-1083532-1-1.html
| 庚 | 辛 | 壬 | 癸 | 甲 | 乙 | 丙 | 丁 | 戊 | 己 | ||
| 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | ||
| 申 | 酉 | 戌 | 亥 | 子 | 丑 | 寅 | 卯 | 辰 | 巳 | 午 | 未 |
| 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 |
'数组怎么读表，是二维表？快速读？
'怎么读为字典呢？
Function YC1(Year&)
If Year < 0 Then Year = Year + 1 '公元前年份+1调整
YC = Mid("庚辛壬癸甲乙丙丁戊己", Year Mod 10 + IIf(Year Mod 10 < 0, 11, 1), 1) & _
Mid("申酉戌亥子丑寅卯辰巳午未", Year Mod 12 + IIf(Year Mod 12 < 0, 13, 1), 1)
End Function
Function YC3(Year&) '皇帝元年公元前2697年
Year = Year + 2696
If Year < 0 Then Year = Year + 1
YC = Mid("甲乙丙丁戊己庚辛壬癸", (Year Mod 10) + 1, 1) _
& Mid("子丑寅卯辰巳午未申酉戌亥", (Year Mod 12) + 1, 1)
End Function
Function YC2(Year&) '皇帝元年公元前2697年
If Year = 0 Then YC = "": Exit Function
If Year < 0 Then Year = 56641 + Year
YC = Mid("庚辛壬癸甲乙丙丁戊己", (Year Mod 10) + 1, 1) _
& Mid("申酉戌亥子丑寅卯辰巳午未", (Year Mod 12) + 1, 1)
End Function
这个应该也好用
10天干：庚辛壬癸甲乙丙丁戊己
对其编码：0 1 2 3 4 5 6 7 8 9
12地支： 申酉戌亥子丑寅卯辰巳午未
对其编码：0 1 2 3 4 5 6 7 8 9 10 12
如2014年，取各位数字4，则天干对应编码为甲。
用12去除2014得余数10，则地支对应编码为午。即，2014为甲午年。 这个对的 while 1==1: 这个错误在于，没理解，干支是60一甲子，并不是天干地支内外层循环，而应该是咬齿轮式的循环 list_tiangan=["甲", "乙", "丙", "丁", "戊", "己", "庚", "辛", "壬", "癸"] #include int n = year - 1984; while(n<0) n += 60; printf("%s%s\n", x[n%10], y[n%12]); int main(int argc, char* argv[])
char TianGan[]={"庚辛壬癸甲乙丙丁戊己"};
char DiZhi[]={"申酉戌亥子丑寅卯辰巳午未"};
int year,tg,dz;
cout<<"请输入年份"<
tg=year%10;
dz=year%12;
string s1,s2;
s1.assign(TianGan,tg*2,2);
s2.assign(DiZhi,dz*2,2);
cout<y=int(input("请输入公历年份（公元前请加“-”）："))t=(y-4)%60%10d=(y-4)%60%12T=["甲","乙","丙","丁","戊","己","庚","辛","壬","癸"]D=["子","丑","寅","卯","辰","巳","午","未","申","酉","戌","亥"]print("{}年为：农历{}{}年".format(y,T[t],D[d]))这个python的并不好用，可能是因为year在vba是保留字
tiangan=['甲','乙','丙','丁','戊','己','庚','辛','壬','癸']
dizhi=['子','丑','寅','卯','辰','巳','午','未','申','酉','戌','亥']
year=int(input("请输入你要查询的年份："))
t=(year%10)-4
d=(year%12)-4
print('\n'+'%d年是：'%year+'\n'+tiangan[t]+dizhi[d]+'年'+'\n')发现好多错误代码
list_dizhi=["子", "丑", "寅", "卯", "辰", "巳", "午", "未", "申", "酉", "戌", "亥"]
list_tiandi=[]
for str in list_tiangan:
for str2 in list_dizhi:
list_tiandi.append(str+str2)
print(list_tiandi)
void f(int year)
{
char* x[] = {"甲","乙","丙","丁","戊","己","庚","辛","壬","癸"};
char* y[] = {"子","丑","寅","卯","辰","巳","午","未","申","酉","戌","亥"};
}
{
f(1911);
f(1970);
f(2012);
return 0;
}