除法

任何数都可以表示表示成二进制，假设商是q，则q可以表示成q=2^k1 + 2^k2 + ... + 2^kn (设k1 > k2 > ... > kn) = a/b，故 a = bq=b（2^k1 + 2^k2 + ... + 2^kn）。改进试商每次让a - b<<k，直到a - b << k > 0且a - b << (k+1) < 0，此时就找到了k1。令a' = a - b <<k = a - b*2^k, 那么q’ = a'/b = （ a - b*2^k）/b = a/b - 2^k = q - 2^k，故q = q' + 2^k。重复以上过程即可找到k2, k3, ... kn，最终算出q。

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int divide(int dividend, int divisor) {     bool negative = ((dividend ^ divisor) >> 31) & 0x1 == 1;     unsigned int a = dividend < 0 ? -dividend : dividend;     unsigned int b = divisor < 0 ? -divisor : divisor;     unsigned int quotient = 0;       while (a >= b)     {         unsigned int k = 0;         for (; a >= b << k && b << k <= (0x80000000 >> 1); ++k);         quotient |= 1 << (a < b << k ? --k : k);         a -= b << k;     }       return negative ? 0-quotient : quotient > INT_MAX ? INT_MAX : quotient; }