sumOfDigits(i) - simply sums all the digits and returns the value.
sortDigits() - takes a number and return the lowest possible number
possible with digits of param passed.
flag = false;
for(int i=n+1, i < 1000; i++) //mention to go upto 1000 in problem
{
if(sumOfDigits(i) == sumOfDigits(n))
{
//a candidate
int sorted i = sortDigits(i);
int sorted n = sortDigits(n);
if(i == n)
{
//we found one
cout<<"\nDesired number is "<<i<<endl; //may abort or continue to find
more
flag = true;
}
}
if(!flag)
cout<<"\nNo such number found"<<endl;
}
On Oct 9, 5:04 pm, wujin chen <wujinchen...@gmail.com> wrote:
> @Aamir , yes, for some N, it will be no ans, then return -1.
>
> 2011/10/9 Aamir Khan <ak4u2...@gmail.com>
>
>
>
>
>
>
>
> > Answer won't be possible in for each N. What would be answer for N=999 ?
>
> > On Sun, Oct 9, 2011 at 4:22 PM, Ankur Garg <ankurga...@gmail.com> wrote:
>
> >> Is it sum of bits or sum of digits ?
>
> >> On Sun, Oct 9, 2011 at 1:39 PM, wujin chen <wujinchen...@gmail.com>wrote:
>
> >>> Given a positive number N, find a minimum number M greater than N, M has
> >>> the same length with N and the sum of the bits are equal.
>
> >>> example:
> >>> N=134 , M=143, // 1+3+4=1+4+3
> >>> N=020, M = 101, //2=1+1
>
> >>> the length of N is less than 1000.
>
> >>> --
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> > --
> > Aamir Khan | 3rd Year | Computer Science & Engineering | IIT Roorkee
>
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