谁在Excel电子表格中欠谁钱algorithm？
我有一群人在一段时间内有不同的开支。 这个时期过后，每个人都有平衡。 在excel中看起来像这样：
人A：6
人B：10
人C：-7,5
人D：-8,5
这段时间之后将会进行和解。 我目前通过手动来做到这一点。 这导致：
C人向A人支付6
C人向B人支付1.5
人D向B人支付8,5。
人A从C获得6
B人得到1.5个人C.
庇隆B从人D获得8,5。
（当涉及更多的人时，可以有多种解决scheme。）
问题是我必须将这个程序应用于一大群人。 所以我的问题是：'如何通过使用excel电子表格algorithm或macros来应用这个谁'欠谁'程序？
我也做了一个excel的版本，但在Open办公室。 你可以下载吗？ 下面的macros可以自己工作。 如果它不应该是小事。 它在OOO中正常工作，并保存为Excel 97/2000工作簿文档。
'this might not be needed in Microsoft Excel, comment it out Option VBASupport 1 'OWES Option Explicit 'Cells(row, col), Private Sub cmd1_Click() 'data is same sheet, from row 4, column 4 'row 4 has names in columns, so 4,4 has name 1, 4,5 has name 2 'row 5 has amounts spent like 6, -10 'output is in columns 3 and 5 dim i dim j,s as String, sum1 s="" 'get number of cells used in row 4 and check if corresponding row 6 column has a number value too i = 4 sum1=0 do while(cells(4,i).Value <> "" and i < 500) j = CDbl(cells(5,i).Value) sum1 = sum1 + j if j <> cells(5,i).Value then MsgBox "Col " & i & " not a number?" End end if i=i+1 loop if i > 499 then Msgbox "too many cols" End end if If sum1 > 0.3 or sum1 < -0.3 then Msgbox "Sum is not near 0 :" & sum1 End End if Dim colCnt as Integer colCnt = i - 4 cells (7,1).Value = "Col count = " & colCnt Dim spent(colCnt) as Double Dim owes1(colCnt ) as String Dim owes2(colCnt ) as String for i= 4 to colCnt + 3 spent(i - 3) = CDbl(cells(5,i).Value) Next Dim cnt,lastNeg, abs1,maxPay ' safety var for never ending loops, only if data bad like many cols and more than .1 diffs lastNeg = 4 dim lastPay1 lastPay1 = 10 dim ii,jj,c1,c2,toPay toPay = 0 On Local Error Goto errh for i= 4 to colCnt + 3 cnt = 0 ii = i - 3 c1 = spent(ii) 'Cells(6,i) = "ok " if spent(ii) > 0.1 and cnt < colCnt Then '//has to take cnt = cnt + 1 for j = lastNeg to colCnt + 3 ' ; j < people.length && spent(ii) > 0.1; j++) jj = j - 3 's = s & Me.Cells(ii,j) & " " if spent(ii) > 0.1 then if spent(jj) < -0.1 Then ' //has to give and has balance to give c1 = spent(ii) c2 = spent(jj) lastNeg = j abs1 = spent(jj) * -1'//can use absolute fn maxPay = abs1 if(maxPay > spent(ii)) Then toPay = spent(ii)' else toPay = abs1 End if spent(ii) = spent(ii) - toPay spent(jj) = spent(jj) + toPay Cells(lastPay1, 3).Value = Cells(4 , j) & " pays " & toPay & " to " & Cells(4 , i ) Cells(lastPay1, 5).Value = Cells(4 , i) & " gets " & toPay & " from " & Cells(4 , j) lastPay1 = lastPay1 + 1 End if End if Next End if Next Msgbox "Done" err.Clear if err.Number <> 0 Then errH: dim yy yy = msgBox("err " & err.Number & " " & err.Description & " Continue", 2) if yy = vbYes Then Resume Next End IF End IF End Sub 谁对谁负责配对有多重要？ 我要问的原因 – 很容易计算出每个人的总成本，然后确定谁欠了钱，谁需要退款。 如果一个人可以成为“银行家”，他可以收回所有的钱，并支付所有的退款。
更简单的问题，如果你有人愿意做银行家。
试图将所有东西配对，很可能不会得到确切的答案，或者可能需要一个或多个人向多个人付款 – 以及一个或多个试图从多于一个人收集的人。
见http://sel2in.com/pages/prog/html/owes.html
Javascript fn
<form> Paste tab seperated list of people on first line, Second line has blanace - positive means they spent more than others, negative means other spent on them (they owe) <br> <textarea id=t1 rows=3 cols=70></textarea> </form> <button onclick=calcOwes() >Calc owes</button> <br> <b>Result:</b> <div id=result>Will appear here if data is good</div> <br> <b>Parsed data for debug:</b> <div id=data1>Will appear here if data is good</div> <br> <hr> <script> function calcOwes(){ /** 2013 Tushar Kapila If Person A: 6 Person B: 10 Person C: -7,5 Person D: -8,5 Then Person C pays 6 to person A. Person C pays 1,5 to person B. Person D pays 8,5 to person B. */ s = document.getElementById("t1").value var line = s.split("\n") //v = confirm("Your Data looks okay?:\n" + s) //if(!v){ return;} if(s.length < 2 || s.indexOf("\n") < 0){ alert("No line sep ") return } people = line[0].split("\t") spent = line[1].split("\t") spent2 = line[1].split("\t") if(spent.length < 2){ alert("Bad data, no spent data " + spent.length + "; 0 " + spent[0] + "; 1 " + + spent[1]) return } if(people.length != spent.length){ alert("People and amounts do not tally. make sure no tabs inside names, spaces are okay") return } sum = 0; data1o = document.getElementById("data1") data1o.innerHTML = ""; for(i = 0;i < people.length; i++){ spent[i] = spent[i].trim() spent[i] = parseFloat(spent[i]) sum += spent[i] s = (1 + i) + " \"" + people[i] + "\" :" + spent[i] + ";<br>" data1o.innerHTML += s; } if(sum > 0.2 || sum < -0.2){ v = confirm("Sum (" + sum + ")is not zero continue?") if(!v){return;} } lastNeg = 0; payDetails = new Array(); getDetails = new Array(); lastPay = 0; for(i = 0;i < people.length; i++){ cnt = 0; if(spent[i] > 0.1 && cnt < people.length){//has to take cnt++ for(j = lastNeg; j < people.length && spent[i] > 0.1; j++){ if(spent[j] < -0.1){//has to give and has balance to give lastNeg = j; abs1 = spent[j] * -1;//can use absolute fn maxPay = abs1 if(maxPay > spent[i]){ toPay = spent[i]; }else{ toPay = abs1 } spent[i] -= toPay spent[j] += toPay payDetails[lastPay] = people[j] + " pays " + toPay + " to " + people[i] getDetails[lastPay] = people[i] + " gets " + toPay + " from " + people[j] lastPay++; } } } } s = "" s2 = "" for(i = 0;i < lastPay; i++){ s = s + payDetails[i] + "<br>" s2 = s2 + getDetails[i] + "<br>" } document.getElementById("result").innerHTML = s + "<br>" + s2 } </script> <br>Sample input 1 (tabs there?) <br><pre> abcd 6 10 -7.5 -8.5 </pre> <br>Sample input 2 <pre> Anna Dan Bobby Scareface Colly Doc Egg face -6 10 -7.3 -8.33 11.67 </pre>