For big number mathematics, the multiplication time complexity is not
O(1).
And the programming complexity is to be considered as well.
On Mar 24, 3:36 pm, radha krishnan <radhakrishnance...@gmail.com>
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 <ak4u2...@gmail.com> 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 <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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