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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