1 #include<iostream>
2 #include<assert.h>
3 using namespace std;
4
5 // 位运算实现加减乘除
6
7 int myAdd(int num1, int num2)
8 {
9 if (num2 == 0) return num1;
10 int sum = 0, carry = 0;
11 sum = num1 ^ num2; // 按位抑或
12 carry = (num1 & num2) << 1;
13 return myAdd(sum, carry);
14 }
15
16 int myAddIter(int num1, int num2)
17 {
18 int ret = 0;
19 while (num2)
20 {
21 ret = num1 ^ num2;
22 num2 = (num1 & num2) << 1;
23 num1 = ret;
24 }
25 return ret;
26 }
27
28 int myNeg(int num1)
29 {
30 // 正数到负数的补码表示： 取反加1
31 return myAdd(~num1, 1);
32 }
33
34 int myMinus(int num1, int num2)
35 {
36 return myAdd(num1, myNeg(num2));
37 }
38
39 // (2^0)*k0 + (2^1)*k1 + (2 ^2)*k2 + ... + (2^31)*k31
40 int myPosMulti(int num1, int num2)
41 {
42 int ret = 0;
43 while (num2)
44 {
45 if (num2 & 1) // 对应当前num2的最低位是1
46 ret = myAdd(ret, num1);
47 num1 <<= 1; // num1 = num1 * 2
48 num2 >>= 1; // num2的当前位用过了就右移
49 }
50 return ret;
51 }
52
53 int myMultiply(int num1, int num2)
54 {
55 if (num1 >= 0 && num2 >= 0)
56 return myPosMulti(num1, num2);
57 if (num1 < 0 && num2 >= 0)
58 return myNeg(myPosMulti(myNeg(num1), num2));
59 if (num1 >= 0 && num2 < 0)
60 return myNeg(myPosMulti(num1, myNeg(num2)));
61 if (num1 < 0 && num2 < 0)
62 return myPosMulti(myNeg(num1), myNeg(num2));
63 }
64
65 // 除法就是由乘法的过程逆推，依次减掉（如果x够减的）y^(2^31),y^(2^30),...y^8,y^4,y^2,y^1。减掉相应数量的y就在结果加上相应的数量。
66 int myPosDiv(int num1, int num2)
67 {
68 num1 = num1 > 0 ? num1 : myNeg(num1);
69 num2 = num2 > 0 ? num2 : myNeg(num2);
70 if (num1 < num2)
71 return 0;
72 long long op = num2;
73 int ret = 0;
74 int flag;
75 int curBit = 0;
76 assert(num2 != 0);
77 while (op <= num1)
78 {
79 op <<= 1;
80 ++curBit;
81 }
82 op >>= 1;
83 --curBit;
84
85 while (op >= num2)
86 {
87 if (num1 >= op)
88 {
89 num1 -= op;
90 ret += 1 << curBit;
91 }
92 op >>= 1;
93 --curBit;
94 }
95 return ret;
96 }
97
98 int myPosDivByMinus(int num1, int num2)
99 {
100 int ret = 0;
101 while (num1 >= num2)
102 {
103 num1 = myMinus(num1, num2);
104 ret = myAdd(ret, 1);
105 }
106 return ret;
107 }
108
109 int myPosComp(int num1, int num2)
110 {
111 bool isZero(int);
112 int temp = 1;
113 num2 = num1 ^ num2;
114 if (isZero(num2)) return 0;
115 while (num2 >>= 1)
116 temp <<= 1;
117 return temp & num1;
118 }
119
120
121 int myDivide(int num1, int num2)
122 {
123 if ((num1 > 0 && num2 > 0) || (num1 < 0 && num2 < 0))
124 return myPosDiv(num1, num2);
125 else
126 return myNeg(myPosDiv(num1,num2));
127 }
128
129 bool isNeg(int num)
130 {
131 return num & 0x8000;
132 }
133
134 bool isZero(int num)
135 {
136 return !(num & 0xFFFF);
137 }
138
139 bool isPos(int num)
140 {
141 return !(num & 0x8000) && (num & 0xFFFF);
142 }
143
144 //int myMod(int a, int b)
145 //{
146 //
147 //}
148
149 int divide(int dividend, int divisor) {
150 int ret = 0;
151 long long curDividend, curDivisor, op;
152 bool flag;
153 int curBit = 0; // 记录商的1在哪一位
154 if (divisor == 0) return INT_MAX; // 往负数方向溢出咋整
155 if ((dividend < 0 && divisor > 0) || (dividend > 0 && divisor < 0))
156 flag = 0; // 负数
157 else
158 flag = 1;
159
160 curDividend = abs(dividend);
161 curDivisor = abs(divisor);
162
163 if (curDividend < curDivisor) return 0;
164
165 op = curDivisor;
166 while (op <= curDividend)
167 {
168 ++curBit;
169 op = op << 1;
170 }
171 op = op >> 1;
172 --curBit;
173
174 while (op >= curDivisor)
175 {
176 if (curDividend >= op)
177 {
178 curDividend -= op;
179 ret += 1 << curBit;
180 }
181 op = op >> 1;
182 --curBit;
183 }
184
185 if (flag)
186 return ret;
187 return -ret;
188 }
189
190 int main()
191 {
192 //INT_MIN -2147483648 INT_MIN 2147483647
193 int a = INT_MAX; //
194 int b = -1;
195 cout << "myAdd: " << myAdd(a, b) << endl;
196 cout << "myMinus: " << myMinus(a, b) << endl;
197 cout << "myMultiply: " << myMultiply(a, b) << endl;
198 cout << "myDivide: " << myDivide(a, b) << endl;
199 // -1010369383, -2147483648
200 cout << "Leetcode: " << divide(-1010369383, INT_MIN) << endl;
201 //cout << "myMod: " << myMod(a, b) << endl;
202 system("pause");
203 return 1;
204 }
