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## 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'); printf("Answer: %d\n", carry); }

2 power 1000 is:

10715086071862673209484250490600018105614048117055336074437503883703510511249361224931983788156958581275946729175531468251871452856923140435984577574698574803934567774824230985421074605062371141877954182153046474983581941267398767559165543946077062914571196477686542167660429831652624386837205668069376

Answer: 1366

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

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Categories: Programming
algorithm, c, C Puzzles

I’m so glad to have stumbled upon your site, it has some pretty awesome stuff. On the subject of Project Euler, especially this problem, I would like to point you to my mathematical/programming blog:

http://www.chaitanyavarier.com

As I have published a post about a Java Big Arbitrary Precision Calculator I recently developed, which might be of interest to you. I will also be posting my own solutions to many PE problems that I’ve solved.