Question

How to sum the digits of the number 2^N in C

Answer 1

#include <stdio.h>
#include <string.h>

#define MAX_DIGITS 4096  // Enough to hold 2^1000

void calculate2PowerNAsString(int N, char result[]) {
    // Calculate 2^N as a string for large numbers
    result[0] = '1';
    result[1] = '\0';
    int len = 1;

    for (int i = 0; i < N; i++) {
        int carry = 0;
        for (int j = 0; j < len; j++) {
            int num = (result[j] - '0') * 2 + carry;
            result[j] = (num % 10) + '0';
            carry = num / 10;
        }
        while (carry > 0) {
            result[len++] = (carry % 10) + '0';
            carry /= 10;
        }
        result[len] = '\0';
    }
}

int sumOfDigits(int N) {
    char result[MAX_DIGITS];
    calculate2PowerNAsString(N, result);

    int sum = 0;
    for (int i = 0; result[i] != '\0'; ++i) {
        sum += (result[i] - '0');
    }

    return sum;
}

int main() {
    int N;

    N = 15;
    printf("Sum of digits of 2^%d is: %d\n", N, sumOfDigits(N));

    N = 100;
    printf("Sum of digits of 2^%d is: %d\n", N, sumOfDigits(N));

    N = 1000;
    printf("Sum of digits of 2^%d is: %d\n", N, sumOfDigits(N));

    return 0;
}



/*
run:

Sum of digits of 2^15 is: 26
Sum of digits of 2^100 is: 115
Sum of digits of 2^1000 is: 1366

*/

 

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