How to solve 100^1000 in C? [closed] - c

Question

Closed. This question needs details or clarity. It is not currently accepting answers.
Want to improve this question? Add details and clarify the problem by editing this post.
Closed 8 years ago.
Improve this question
This Question is from Hacker Earth website as follow
Input: First line contains t, the number of test cases. After that t lines follow each containing a pair of space separated integers i.e. a and b.
Output: For each test case, you need to print the sum of digits in a^b.
Constraints:
1<=t<=20000
1<=a<=100
1<=b<=1000
Sample Input (Plaintext Link)
5
2 10
3 3
5 2
2 7
100 1000
Sample Output (Plaintext Link)
7
9
7
11
1
Explanation: Consider the first test case.
2^10 = 1024
Now, 1 + 0 + 2 + 4 = 7
So, output for this test case is 7.
Now I have solve this problem as follows:
#include<stdio.h>
#define test_max 20000
#define a_max 100
#define b_max 10000
long int power(int,int);
long int check_num(int,int);
int main()
{
int t;
long int i,rem,j=0,sum=0;
long int pr;
scanf("%d",&t);
int a[t],b[t];
for(i=0;i<t;i++)
{
scanf("%d",&a[i]);
scanf("%d",&b[i]);
if(a[i]>a_max||a[i]<=0||b[i]>b_max||b[i]<=0)
{
// printf("Please Check the Enter number limitation");
exit(0);
}
}
for(i=0;i<t;i++)
{
int check=0;
check=check_num(a[i],b[i]);
if(check==0)
{
pr=power(a[i],b[i]);
//printf("%d\t%d\t%ld\t",a[i],b[i],pr);
while(pr!=0)
{
rem=pr%10;
sum=sum+rem;
pr=pr/10;
}
printf("%d\n",sum);
sum=0;
}
else if(check==1)
{
printf("1\n");
}
}
return 0;
}
long int check_num(int a,int b)
{
if(a==100||a==10||a==1) return 1;
else return 0;
}
long int power(int n,int p)
{
int i;
long int num=n;
for(i=1;i<p;i++)
{
num=num*n;
}
return num;
}
But it's not a perfect answer. Can anyone tell me how to solve this problem?
The link of this problem is here: HERE
and I know the 100^1000 is not possible without using any library and I just have to avoid this but how?

Answer 1

100^1000 is too large a number to fit into a long, or any other basic number type.
You need to pick a big integer (also called big number) library that helps you manage arbirtrarily big numbers.
Try libGMP, it's widely used.

Answer 2

Generally speaking, When you talk about large numbers, you have 3 main options:
1) Use a library that manipulates big numbers
2) Implement your own functions for big numbers (use arrays to store the numbers - the length will be equal to the number of digits)
3) Find a Math approach
I will add some details for the 3rd approach, because the 1st and the 2nd are not very complicated.
If you need only the sum for those digits, try to avoid calculating that big number.
Think about logarithms and their properties when you work with large numbers.
For example, your particular case can be written in the following form:
100^1000 = (10^2)^1000 = 10^(2*1000) * 1 = 10^2000 + 0 = 10^2000 * 1
You know that the first part (10^2000) only adds some zeros => it doesn't influence your result (sum). So, you can try to split that big numbers in 2 parts ( a part that doesn't influence the sum and a useful part for the sum - 1 ).
How can you do this using logarithms for a more general problem? The following observation can generalize the things that were mentioned above:
a^x = 10^lg(a^x) = 10^(x*lg(a))
where lg = log in base 10
Then you can write your number as: 10^c * k or 10^c + k or 10^c - k, etc. depending on your numbers.
This explanation is just for guidance, to give you a taste. But the idea is the same: think about a mathematical solution based on some properties before you try other approaches.