Sameer Kumar Boxi

• Apr 27th, 2006

 

int computeXn(int x, int n)

{

  if(n == 2)

  {

    return x*x;

  }

 else if(n % 2 == 0)

 {

  int y = computeXn(x, n/2);

  return y*y;

 }

 else if(n % 2 == 1)

 {

  int y = computeXn(x, n/2);

  return y*y*x;

  }

}

Vinay

• Apr 28th, 2006

 

The program has a bug not all the control path returns a value

deepak kumar

• May 3rd, 2006

 

#include#includelong int a;long int func(long int x,long int n);int s=1;void main(){ long int x,n,p=1; scanf("%li",&x); for (n=2;n<15;n++) { a=x; s=1; if (n%2==1) p=func(x,n-1)*x; else p=func(x,n); printf("\n%li %li = %li",x,n,p*s); } getch();}long int func(long int x,long int n){ if (n==1) return x; if (n%2==1) {

mkag

• May 11th, 2006

 

int Sq(int x, int n) {

    int y = 0;

 

    if (n < 0) return -1;

    if (n == 0) return 1;

    if (n == 1) return x;

 

    if (n % 2 == 0) {

        y = Sq(x, n/2);

        return y * y;

    }

 

    y = Sq(x, n/2);

    return x * y * y;

}

Ashish Bhawsar

• May 13th, 2006

 

long Pow(int x, int n){ long Power=1,z=x,m=n; while(m>0) { while(m%2==0) { m=m/2; //Integer division z=z*z; } m--; Power*=z; } return Power;}

hari

• Aug 20th, 2006

 

int xpwrn(int x,int n){ if(n==1) return x ; else if(n>1) { if(n%2==0) { xpwrn(x,n/2)*xpwrn(x,n/2); } else { x*xpwrn(x,n/2)*xpwrn(x,n/2); } }}

jinhu

• Aug 8th, 2007

 

my anwser:
int xpwrn(int x,int n)
{
    int i=0;
    int rv=1;
    int n=x;
    for(;i<32&&(1<<i)<x;++i)
    {
            if((1<<i)&x != 0)
            {
               rv *= n;
             }
             n  *=x;
    }
    return rv;
}

Sandeep Phukan

• Nov 7th, 2007

 

class powerRecurse {
 
 
 
 public int powerRec(int x,int n){
  
  if ( n==1 ) return x;
  else {
   return x*powerRec(x,--n);
  }
  
 }
 
 
 
 public static void main(String[] a){
  
  System.out.println(new powerRecurse().powerRec(2,4));
 }
 
 
 
}

rahuldbhagat

 Profile Answers by rahuldbhagat 

• May 3rd, 2008

 

Here is the algorithm that works well and takes O(log n) time.

Algorithm Exponientiate(x,n)
//computes x^n  for an integer n>=0
{ 
    m:= n;
    power:=1;
    z:=x;
    while( ( m mod 2 )== 0 ) do
            {
                  m:= m/2 ; //take the floor of the result.
                  z:=z^2;  //square of z
              
            
             }
     m:=m-1;
    power:=power*z;


 }
 return power;
}

ashgoel

 Profile Answers by ashgoel 

• Jun 24th, 2010

 

represent N=b(k)b(k-1)....b(1)b(0) as bit representation

then N=summation i=o to k   b(i)*2^i

so x can be represented as 

x(N)=x(b0)*x(2b1)*x(4b2)...*x(2(k)bk)

bois nothing but Nmod 2

long power(int x, int n)

{

long power=1;

long y=x;

while (n>0)

while (n%2 ==0)

{

y*=y;

n/=2;

}

n-=1;

power *=y;

}

return power;

}

}

}