Just a MINOR change.
please change the RED line(I've marked it below) in your code by the
following line:
*return (x*((z*z)%MOD))%MOD;*
and enjoy an *ACCEPTED *verdict.*
*
*Wasif Hossain**
*
B.Sc. Student of Final Semester,
Computer Science and Engineering(CSE),
Bangladesh University of
- Take another sum array which contains sums of original array at each
index, here sum[0] = a[0]; sum[1] = a[0] + a[1]
;...sum[i]=a[0]+a[1]+...a[i];
- Traverse the sum array and search for duplicates.
ex: a[] = {1,2,-4,-3, 6 ,-3};
s[] = { 1,3,-1,-4,2,-1 };
In the sum array we have
index = 0 1 2 3 4 5 6
ar = 0 1 2 -4 -3 6 -3
sumar = 0 1 3 -1 -4 2 -1
first index where we get the number which has already appeared in sumar will
be the last index of sub array whose sum will be zero and (index of first
apperance of this number + 1) in sumar will be the start index.
@Ravindra
To check the particular number square can be written as sum of two squares
or not.
If it has any prime factor x such that x % 4 == 1 then only.
Now about time complexity.
suppose u have given array is.
10 6 13 9 17 4 18 12 1 5.
now u can easily skip the numbers(1, 4, 6, 9, 12, 18);
+1 for Sumanth's solution.
On Oct 20, 12:03 pm, anshu mishra anshumishra6...@gmail.com wrote:
index = 0 1 2 3 4 5 6
ar = 0 1 2 -4 -3 6 -3
sumar = 0 1 3 -1 -4 2 -1
first index where we get the number which has already appeared in sumar will
be the last index of sub array whose
Excerpt from The C Programming Language by Kerninghan Ritchie -
*int fflush(FILE *stream) - *On an output stream, fflush causes any
buffered but unwritten data to be written; *on an input stream, the effect
is undefined*.
So fflush was never meant for stdin.
- Ravindra
On Sun, Oct 9, 2011 at
