Multiply By seven [closed]

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I was asked this interview question: what is the quickest way to multiply a number by 7. she told me not to use any of the + , - , * , / operators. In tense,i could not answer the question.

I know the quickest way to multiply a number by 8 is n<<3 but can n*7 be acheived?

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Assuming your compiler isn't terrible, n*7.

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Meets almost all your requirements :)

#include <stdio.h>

int mult7(int i)
{
    int res;
    __asm__("imull  $7, %1, %0" : "=r" (res) : "r" (i));
    return res;
}

int main()
{
    printf("%d", mult7(12)); //output: 84
    return 0;
}

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No forbidden operators (+, -, *, or /) used :)

/* multiply by 7 without using the forbidden operators */
/* if `n` is larger than `INT_MAX / 7`
** or smaller than `INT_MIN / 7`
** the behaviour is undefined. */
int mult7(int n) {
  int n7 = n;
  n7 *= 7;
  return n7;
}

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n*7 == (n<<3) - n

Not sure if that will be better then normal multiplication by 7 though

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The correct answer is a combination of

a) n*7

because this is exactly the kind of micro-optimization that the compiler is perfectly capable of figuring out on its own.

b) the interviewer has a poor understanding of modern optimizing compilers

Alternatively, if the interviewer isn't clueless, an answer different than a) above suggests that the interviewee does not understand optimizing compilers and is thus a poor candidate for the job. :-/

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The best answer I can come up with is to write "+" operation using while loop and bitwise operators per each bit. Something like this:

int add(int x, int y) {
  int a, b;
  do {
    a = x & y; b = x ^ y; x = a << 1; y = b;
  }while(a);
  return b;
}

And then sum up n, n << 1, n << 2.

"+" is easy to write with while loop, if you can come up with way to write "-" with while loop (which i am not sure how to do exactly) than you can do (n << 3) - n.

source:http://geeki.wordpress.com/2007/12/12/adding-two-numbers-with-bitwise-and-shift-operators/

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It's likely either

n*7

or

(n<<3) - n

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Maybe this is a valid answer as well : pow(10, log10(n) + log10(7))

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oh you were so close!

= (n << 3 - n)

= n * 8 - n

= n(8 - 1)

= n * 7

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Assuming that the interviewer is looking for a shift-and-add kinda answer, I guess you could use:

int ans = (num<<2) + (num<<1) + num;

I guess this a crude way of testing if the interviewee knows of this particular algo.

---

int main()
   {
        int i;
        int x;
        int n;
        x = n;
        for (i = 2; i < 8; i++)
        {
             n += x;
        }
        return 0;
    }

EDIT: converted to c (i think) and realized there is no .= in c so had to switch to +=. Still not the same as + technically but a bit questionable.

I know it uses the ++ but that certainly not the same as "+" and .= avoids the stipulation.