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新概念招聘3.0

如何不用+、-实现加法和减法?

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谁知道底层实现的算法?

 

1 个答复

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#include <stdlib.h> /* atoi() */
#include <stdio.h>  /* (f)printf */
#include <assert.h> /* assert() */

int add(int x, int y) {
    int carry = 0;
    int result = 0;
    int i;

    for(i = 0; i < 32; ++i) {
        int a = (x >> i) & 1;
        int b = (y >> i) & 1;
        result |= ((a ^ b) ^ carry) << i;
        carry = (a & b) | (b & carry) | (carry & a);
    }

    return result;
}

int negate(int x) {
    return add(~x, 1);
}

int subtract(int x, int y) {
    return add(x, negate(y));
}

int is_even(int n) {
    return !(n & 1);
}

int divide_by_two(int n) {
    return n >> 1;
}

int multiply_by_two(int n) {
    return n << 1;
}

int multiply(int x, int y) {
    int result = 0;

    if(x < 0 && y < 0) {
        return multiply(negate(x), negate(y));
    }

    if(x >= 0 && y < 0) {
        return multiply(y, x);
    }

    while(y > 0) {
        if(is_even(y)) {
                x = multiply_by_two(x);
                y = divide_by_two(y);
        } else {
                result = add(result, x);
                y = add(y, -1);
        }
    }

    return result;
}

int main(int argc, char **argv) {
    int from = -100, to = 100;
    int i, j;

    for(i = from; i <= to; ++i) {
        assert(0 - i == negate(i));
        assert(((i % 2) == 0) == is_even(i));
        assert(i * 2 == multiply_by_two(i));
        if(is_even(i)) {
                assert(i / 2 == divide_by_two(i));
        }
    }

    for(i = from; i <= to; ++i) {
        for(j = from; j <= to; ++j) {
                assert(i + j == add(i, j));
                assert(i - j == subtract(i, j));
                assert(i * j == multiply(i, j));
        }
    }

    return 0;
}
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