On Sunday, December 23, 2012 9:07:53 PM UTC+5:30, Anil Sharma wrote: > > please suggest some efficient solution to check perfect square condition . > no matter how much large number is... eg..i/p-8949 o/p-93 > there is no specific algorithm for it but yeah you can use binary search > for (bisection method )... or use the property that number of divisor of a > perfect square is always odd >
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