Inconsummate numbers in base 10: Difference between revisions
Inconsummate numbers in base 10 in FreeBASIC
(→{{header|Free Pascal}}: Faster version checking Digitalsum, testing for max used factor.) |
(Inconsummate numbers in base 10 in FreeBASIC) |
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Inconsummate number 100000: 536081
</pre>
=={{header|FreeBASIC}}==
{{trans|Action!}}
<syntaxhighlight lang="freebasic">Dim As Boolean consummate(0 To 999999)
For j As Integer = Lbound(consummate) To Ubound(consummate)
consummate(j) = False
Next j
consummate(1) = True
Dim As Integer maxSum = 9
Dim As Integer ub = Ubound(consummate)
While ub > 10
maxSum += 9
ub /= 10
Wend
Dim As Integer maxNumber = Ubound(consummate) * maxSum
Dim As Integer tn = 1, hn = 0, th = 0, tt = 0, ht = 0, mi = 0, tm = 0
Dim As Integer sumD, n, d, dRatio
For n = 10 To maxNumber Step 10
sumD = tm + mi + ht + tt + th + hn + tn
For d = n To n + 9
If d Mod sumD = 0 Then ' d is comsummate
dRatio = d / sumD
If dRatio <= Ubound(consummate) Then consummate(dRatio) = True
End If
sumD += 1
Next d
tn += 1
If tn > 9 Then
tn = 0
hn += 1
If hn > 9 Then
hn = 0
th += 1
If th > 9 Then
th = 0
tt += 1
If tt > 9 Then
tt = 0
ht += 1
If ht > 9 Then
ht = 0
mi += 1
If mi > 9 Then
mi = 0
tm += 1
End If
End If
End If
End If
End If
End If
Next n
Dim As Integer i = 0, count = 0
Print "The first 50 inconsummate numbers:"
While i < Ubound(consummate) And count < 100000
i += 1
If Not consummate(i) Then
count += 1
If count < 51 Then
Print Using "######"; i;
If count Mod 10 = 0 Then Print
Elseif count = 1000 Or count = 10000 Or count = 100000 Then
Print Using !"\nInconsummate number ######: ######"; count; i;
End If
End If
Wend
Sleep</syntaxhighlight>
{{out}}
<pre>The first 50 inconsummate numbers:
62 63 65 75 84 95 161 173 195 216
261 266 272 276 326 371 372 377 381 383
386 387 395 411 416 422 426 431 432 438
441 443 461 466 471 476 482 483 486 488
491 492 493 494 497 498 516 521 522 527
Inconsummate number 1000: 6996
Inconsummate number 10000: 59853
Inconsummate number 100000: 536081</pre>
=={{header|J}}==
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