you need to try something better as limits of A and B are very large :)
you can not run a loop from A to B
i have not tried it but the logic is there will be many nos which will give
the same value and we dont need to calculate for them all explicitply :)
On Fri, Jun 17, 2011 at 2:52 PM, KK <kunalkapadi...@gmail.com> wrote:
> To remove all digits left of the rightmost digit one in the binary
> representation of some integer what we need to do is this:
> ans = no & -no
> and this is what is exactly asked in this problem of SPOJ:
> www.spoj.pl/problems/MZVRK/
>
>
> #include<iostream>
> using namespace std;
> int main()
> {
> unsigned long long int a, b, sum;
> while(scanf("%lld%lld", &a, &b) != EOF)
> {
> sum = 0;
> while(a != (b+1))
> {
> sum += (a & -a);
> a++;
> }
> printf("%lld\n", sum);
> }
> return 0;
> }
>
> Its giving TLE on some 10th case...
>
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--
Sunny Aggrawal
B-Tech IV year,CSI
Indian Institute Of Technology,Roorkee
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