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 <shashank7andr...@gmail.com> 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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