Error on Normalized Equation for Josephus Problem

Question

//
//  main.c
//  단기 연구 과제
//  Created by Shiro Nai on 07/05/2019.
//  Copyright © 2019 Shiro Nai. All rights 
//  reserved.


#include <stdio.h>
#include <math.h>

int main(int argc, const char * argv[]) {
    double n;

    printf("How many people?");
    scanf("%lf", &n);

    int k = floor(log10(n)/log10(2));

    printf("The last standing person's number is %lf.", (2*(n-(2^k)))+1);
    return 0;
}

The output is different from what I expected. For example, when n=5 the answer is 3, but I get 11 for the output.

Do you know what `2^k` is? It is using the exclusive-or operator, but I guess you are raising to `2` to the power `k`? — Weather Vane, May 06 '19 at 16:43
^ is not being treated like a power-of operator, but rather an exclusive-or operator, which both do different things. — Frontear, May 06 '19 at 17:03
C has a [log2()](https://en.cppreference.com/w/c/numeric/math/log2) function, btw. — Shawn, May 06 '19 at 17:06
You'll probably also be interested in [exp2()](https://en.cppreference.com/w/c/numeric/math/exp2). — Shawn, May 06 '19 at 17:08
The `floor()` changes things though. Nevermind that attempt at simplification. — Shawn, May 06 '19 at 17:20

score 3 · Answer 1 · answered May 06 '19 at 17:07

3

The ^ symbol is not a power operator. This does not do any kind of math, this is an exclusive-or operator.

In order to do power calculations, you need pow() from math.h

#include <stdio.h>
#include <math.h>

int main() {
    printf("The value of 11 ^ 3 = %.2f", pow(11.0, 3)); // outputs 1331.00
}

Sources:

answered May 06 '19 at 17:07

Frontear

1,150
12
25

1

You could point to the C standard better than some pages one of which says "cplusplus" – alx - recommends codidact May 06 '19 at 19:19

Error on Normalized Equation for Josephus Problem

1 Answers1