9、編程實現兩個正整數的除法

/************************************************************************//* 9、編程實現兩個正整數的除法                                            *//************************************************************************/// abs(x)/abs(y) = h, 則 (h+1)* abs(y) > abs(x) >= h * y, 利用二分尋找求h //找到滿足條件的k， 時間複雜度o（1）int div1(const int x, const int y){if(y == 0)throw new exception("divided by zero!");if(x == 0)return 0;//保證x、y可以去到0xffffffffunsigned tempX = abs(x);unsigned tempY = abs(y);unsigned h = 1;unsigned hy = tempY;//最多32次while(hy < tempX){hy <<= 1;        //tempY * 2h <<= 1;}//最多31次//result一定在[h/2, h)之間 [l, h)unsigned l = h >> 1;while(l < h - 1){unsigned mid = (l + h) >> 1;           //(l +h) /2unsigned midy = mid * tempY;if(midy == tempX){l = mid;break;}else if(midy < tempX)l = mid;elseh = mid;}//判斷符號if((x > 0 && y < 0) || (x < 0 && y > 0))return -1 * l;elsereturn l;}void testOfDiv(){int x = rand() % 1000000;int y = rand()% 10000000;int ret1= 0;cout << "x/y" << endl;CLOCK{ret1 = x / y;}int ret2 = 0;cout << "div1(x/y)" << endl;CLOCK{ret2 = div1(x, y);}assert(ret1 == ret2);}