You don't need to use BigNum or long int for this program.
Both n & k should be less than 1000.
Since there is no restriction on k,you don't need Bignum
Since both n,k are restricted,you don't need bignum.
if n>5, simply reject the input and return false
On Fri, Jun 8, 2012 at 11:01 AM, Dave <dave_and_da...@juno.com> wrote:
> @victor: But if K <= 1000, then the largest N you have to deal with is 4,
> since 4^4 < 1000 but 5^5 > 1000. So your code looks like this:
>
> int IsNtoNEqualK( int N, int K)
> {
> return (N==1)&&(K==1) || (N==2)&&(K==4) || (N==3)&&{K==27) ||
> (N==4)&&(K==256);
> }
>
>
> On Thursday, June 7, 2012 5:14:00 PM UTC-5, Victor Manuel Grijalva
> Altamirano wrote:
>
>> Hi, everybody!!!
>> I have the follow quest...
>>
>> I have two numbers N and K, i need to check that N^N = K.
>> for example:
>> if N=2 and K=4 , 2^2 = 4 so return true;
>> if N=3 and K=26 , 3^3 != 26 so return false
>> But 0<=N , K<=1000 so N^N could be have 1000 digits.
>>
>> I program in C++, and i can use Bignum (array manipulation) + fast
>> power(binary power) but i want to know if exist a mathematical property.
>>
>>
>> --
>> Victor Manuel Grijalva Altamirano
>> Universidad Tecnologica de La Mixteca
>>
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Abhishek Sharma
Under-Graduate Student,
PEC University of Technology
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