多元一次方程往往採用迴圈求解。筆者在與網友們討論乙個問題（過程中，琢磨出一種演算法，採用遞迴進行多元一次方程的求解。並將解分為整數解和 非負整數解兩種情況，請大家指教。
private sub command1_click() '演示求x1+x2+x3+x4+x5=10整數解
text1.text = ""
dim answer as string
answer = getresult(5, 10, true) '賦值
dim temp
temp = split(answer, vbcrlf)
for i = 0 to ubound(temp)
temp(i) = "解" & i + 1 & "：" & vbtab & temp(i) ' add index
next
answer = join(temp, vbcrlf)
text1.text = "方程 x1+x2+x3+x4+x5=10 共有 " & ubound(temp) + 1 & " 個整數解：" & vbcrlf & answer 'show all answer in textbox
end sub
private sub command2_click() '演示求x1+x2+x3+x4+x5=10非負整數解
text1.text = ""
dim answer as string
answer = getresult(5, 10, false) '賦值
dim temp
temp = split(answer, vbcrlf)
for i = 0 to ubound(temp)
temp(i) = "解" & i + 1 & "：" & vbtab & temp(i) 'add index
next
answer = join(temp, vbcrlf)
text1.text = "方程 x1+x2+x3+x4+x5=10 共有 " & ubound(split(answer, vbcrlf)) + 1 & " 個非零整數解：" & vbcrlf & answer 'show all answer in textbox
end sub
private sub command3_click() '演示無解情況
text1.text = ""
dim answer as string
answer = getresult(5, 3, false)
dim temp
temp = split(answer, vbcrlf)
for i = 0 to ubound(temp)
temp(i) = "解" & i + 1 & "：" & vbtab & temp(i)
next
answer = join(temp, vbcrlf)
text1.text = "方程 x1+x2+x3+x4+x5=3 共有 " & ubound(split(answer, vbcrlf)) + 1 & " 個非零整數解：" & vbcrlf & answer
end sub
'求解函式
function getresult(byval n as integer, byval sum as integer, optional allowzero as boolean = true) as string
dim temp() as string, i as long
if n = 2 then '二元方程
if allowzero = true then
redim temp(sum)
for i = 0 to sum ' allow zero
temp(i) = "x1=" & i & ",x2=" & sum - i
next
getresult = join(temp, vbcrlf)
erase temp
else
redim temp(1 to sum - 1) 'forbid zero
for i = 1 to sum - 1
temp(i) = "x1=" & i & ",x2=" & sum - i
next
getresult = join(temp, vbcrlf)
erase temp
end if
end if
if n > 2 then
if allowzero = true then
redim temp(sum)
for i = sum to 0 step -1 ' allow zero
temp(i) = replace(getresult(n - 1, i, true), vbcrlf, ",x" & n & "=" & sum - i & vbcrlf) & ",x" & n & "=" & sum - i
next
getresult = join(temp, vbcrlf)
erase temp
else
if sum < n then msgbox "無解！": exit function '無解情況
redim temp(1 to sum - n + 1) 'not allow zero
for i = 1 to sum - n + 1
temp(i) = replace(getresult(n - 1, sum - i, false), vbcrlf, ",x" & n & "=" & i & vbcrlf) & ",x" & n & "=" & i '遞迴
next
getresult = join(temp, vbcrlf)
erase temp
end if
end if
end function
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