Yeah i know the algorithm is not very much efficient...But i was trying to
show how to tackle with big integers in C++ (i.e, using arrays)

On Thu, Mar 24, 2011 at 1:06 PM, radha krishnan <
[email protected]> wrote:

> You can do it in (log n) assuming multiplication is O(1)
> suppose u are going to calculate 8 power 33
> u compute 8 power 16 and multiply with the same to get 8 power 32
> then multiply with 8 to get the result
>
> On Thu, Mar 24, 2011 at 1:04 PM, AAMIR KHAN <[email protected]> wrote:
> > Try this...
> > #include <iostream>
> > #include <cmath>
> > using namespace std;
> > #define DIGITS 10001
> > void mult(int N,int pro[],int &len) {
> >    int carry = 0;
> >    for(int i=0;i<len;i++) {
> >       int temp = pro[i]*N + carry;
> >       pro[i] = temp%10;
> >       carry = temp/10;
> >    }
> >
> >    if(carry>0) {
> >       pro[len] = carry;
> >       len++;
> >    }
> >
> > }
> > int main() {
> >    int t,N,E;
> >
> >    scanf("%d",&t);
> >    while(t--) {
> >       int pro[DIGITS];
> >       scanf("%d %d",&N,&E);
> >       if(N==1) {
> >       printf("1 1\n");
> >       continue;
> >       }
> >       pro[0] = 1; int len = 1;
> >       for(int i=0;i<E;i++) {
> >            mult(N,pro,len);
> >       }
> >
> >       for(int i=len-1;i>=0;i--) {
> >          printf("%d",pro[i]);
> >      }
> >       printf("\n");
> >
> >    }
> >    return 0;
> > }
> >
> > Here t stands for number of testcases...
> > N => the number for which power is to be calculated..
> > E => the exponent..
> > On Thu, Mar 24, 2011 at 12:22 PM, bittu <[email protected]>
> wrote:
> >>
> >> How you will print the 100th power of a single digit( which is of type
> >> int). How do you maintain that big number in memory?
> >>
> >>
> >> Lets C The Approach
> >>
> >> Thank & Regards
> >> Shashank
> >>
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