Home > Programming > Project Euler: Problem 16: Find the sum of all the digits in 2 power 1000

## Project Euler: Problem 16: Find the sum of all the digits in 2 power 1000

This problem is simple and straight forward. I’ve used a character buffer to store the powers of 2 for each iteration starting from 1 to 1000.

```#include <stdio.h>
#include <stdlib.h>
#include <string.h>
char *strrev(char *str)
{
char *p1, *p2;

if (! str || ! *str)
return str;
for (p1 = str, p2 = str + strlen(str) - 1; p2 > p1; ++p1, --p2)
{
*p1 ^= *p2;
*p2 ^= *p1;
*p1 ^= *p2;
}
return str;
}
int main(void)
{
char *digits = (char *)malloc(sizeof(char) * 2000);
memset(digits, 0, 2000);
digits[0] = '1';
unsigned int carry = 0;
int i, j;
for (i = 0; i < 1000; i++) {
carry = 0;
for (j = 0; j < strlen(digits); j++) {
int m = (digits[j] - '0') * 2 + carry;
carry = (m / 10) % 10;
digits[j] = (m % 10) + '0';
}
if (carry)
digits[j] = carry + '0';
}
strrev(digits);
printf("2 power 1000 is: %s\n", digits);
carry = 0;
for (int i = 0; i < strlen(digits); i++)
carry += (digits[i] - '0');
}

```

2 power 1000 is:

```10715086071862673209484250490600018105614048117055336074437503883703510511249361224931983788156958581275946729175531468251871452856923140435984577574698574803934567774824230985421074605062371141877954182153046474983581941267398767559165543946077062914571196477686542167660429831652624386837205668069376
```

real 0m0.085s
user 0m0.015s
sys 0m0.062s

One interesting thing to notice here is that the above same code can be written in python with in a single line using comprehensions as follows. 🙂

```
sum([int(n) for n in str(2**1000)])

```

>>> sum([int(n) for n in str(2**1000)])
1366