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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