# algorithm: how to seperate a number into a group of small numbers

1) there are 6 numbers 1,2,4,8,16,32,64. Increase by power 2.
2) given a number N <=64, for example 59, see what numbers in 1) consist of 59.
3) a number can be used only once.
It's better to code in DOS shell.
For example, 59 = 32+16+8+2+1
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Commented:
Good evening.... This was a lovelly little teaser. Please see my solution below:

@echo off
if "%1"=="" exit /b
if %1 leq 0 exit /b

set dec=%1
set result=
set bit=1

:loop1
set /a bit=bit*2
if %bit% leq %dec% goto loop1

set /a bit=bit/2

:loop2
if %bit% leq %dec% (
set /a dec-=bit
if not defined result (
set result=%bit%
) else (
set result=%result%+%bit%
)
)
set /a bit=bit/2
if %bit% gtr 0 goto loop2

echo %1 = %result%
0

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Author Commented:
Excellent. One more question: if the number < 1,200,000,000, it's OK. Do you have any idea when the number is larger than it, say 8,000,000,000?
0
"Batchelor", Developer and EE Topic AdvisorCommented:
This will work only for 32bit signed integer, meaning 2,17 billion.

The algorithm is standard binary coding, because of "power 2" in point 1. Why to obscure that fact? Sounds like homework ...

0
Commented:
Thank you Qlemo....

jl66
There is a limit in DOS when performing integer maths. The solution provided uses mathematical evaluations to arrive at the output.And as you've discovered, this noly works upto a certain limit.

This modified loop1 will handle numbers upto 2,147,483,647.

@echo off

if "%1"=="" exit /b
if %1 leq 0 exit /b

set result=

set dec=%1
set bit=1
set tmp=1

:loop1
set /a tmp=tmp*2
if %tmp% gtr %dec% goto end_loop1
set bit=%tmp%
goto loop1
:end_loop1

:loop2
if %bit% leq %dec% (
set /a dec-=bit
if not defined result (
set result=%bit%
) else (
set result=%result%+%bit%
)
)
set /a bit=bit/2
if %bit% gtr 0 goto loop2

echo %1 = %result%
exit /b
0
Author Commented:
Not homework but simplied real issue.
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