# 二元一次不定方程的大衍求一术解法的VB源码

#### 二元一次不定方程的大衍求一术解法的VB源码

我们清楚，二元一次不定方程可用欧几里德扩展算法，或者同余方程的欧拉解法，以及中国传统的方法大衍求一术来解。同余方程的欧拉解法程序算法复杂，且需经递归。而欧几里德扩展算法如果不递归，则也需要进行回退求解。所以，只有大衍求一术才是最简单的算法。这是08年就写出的老代码了，VB6的源码。从本人其它博客现搬到iteye.com网站,分享给更多的人。以下是VB程序的源码。

Option Explicit

'visual basic source code for finding x, y such that linear diophantine equation

Function gcd(ByVal a As Long, ByVal b As Long)
'get great common division
Dim c As Long, tmp As Long

If (a < b) Then
tmp = a
a = b
b = tmp
End If

Do While (b <> 0)
c = b
b = a Mod b
a = c
Loop
gcd = a

End Function

Function gma(ByVal g As Long, ByVal m As Long) As Long
'get special solution for linear diophantine: ax+by=1
Dim LT As Long, RT As Long, LD As Long, RD As Long
Dim tmp As Long, x As Long

If Abs(g) > Abs(m) Then
g = g Mod m
End If

LT = 1: LD = 0: RT = g: RD = m
x = 0

Do While (RT <> 1)
x = RD \ RT
RD = RD Mod RT
LD = x * LT + LD

x = RT \ RD
If RD = 1 Then
LT = (x - 1) * LD + LT
RT = 1
Exit Do
Else
RT = RT Mod RD
End If
LT = x * LD + LT
Loop
gma = LT

End Function

Function Diophantine(ByVal a As Long, ByVal b As Long, ByVal c As Long, _
x As Long, y As Long, ar As Long, br As Long) As Boolean
'Find the x,y such that a*x + b*y = k*gcd(a,b)
Dim g As Long, tmp As Long, x0 As Long, y0 As Long
Dim am As Long, bm As Long, t As Long

g = gcd(Abs(a), Abs(b))
If (c Mod g <> 0) Then
Diophantine = False: Exit Function
End If
If a < 0 Then
a = -a: b = -b: c = -c
End If
br = a / g: ar = b / g
am = br: bm = ar
If ar > br Then
bm = ar Mod br
End If
If am > bm Then
y0 = gma(Abs(bm), Abs(am))
Else
y0 = 1
End If
If (b < 0) Then
y0 = -y0
End If
x0 = (1 - ar * y0) / br
x = c * x0 / g: y = c * y0 / g
'let y to mininized value
t = y \ br
y = y Mod br
x = x + ar * t

ar = -ar

Diophantine = True

End Function

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