Answer Posted / m.choudhury

The problem is 2^(axxxx) where x belongs to {0,1,.....9} & a
belongs to {1,2,3,.....9}. This is clearly not equivalent to
2*(axxxx).

The solution will be easier if we can give the answer in
HEXADECIMAL format.

2^2=(4)DEC=(100)BINARY=(4)HEX
2^4=(16)DEC=(10000)BINARY=(10)HEX
2^7=(128)DEC=(10000000)BINARY=(80)HEX
.
.
.
2^n=(X)DEC=(100.....0)BINARY{n no. of zero after 1}=(Z)HEX
(X is the decimal of 2^n, Z is HEXADECIMAL of 2^n)

To get Z get HEX of (1a) where a = n%4 is the number of
zeros after 1.

Then path n/4 no. of zeros with that.

can anyone suggest the code for integer representation of
2^n , (where n is any integer), with polynomial time
complexity ?

Is c procedural or object oriented?

953

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#include main() { int *p, *c, i; i = 5; p = (int*) (malloc(sizeof(i))); printf(" %d",*p); *p = 10; printf(" %d %d",i,*p); c = (int*) calloc(2); printf(" %d ",*c); }

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