如何将一个数除3而不使用*、/、+、-、%等运算符?
号码可以有签名,也可以没有签名。
如何将一个数除3而不使用*、/、+、-、%等运算符?
号码可以有签名,也可以没有签名。
当前回答
哪里InputValue是数字除以3
SELECT AVG(NUM)
FROM (SELECT InputValue NUM from sys.dual
UNION ALL SELECT 0 from sys.dual
UNION ALL SELECT 0 from sys.dual) divby3
其他回答
这是经典的2进制除法算法
#include <stdio.h>
#include <stdint.h>
int main()
{
uint32_t mod3[6] = { 0,1,2,0,1,2 };
uint32_t x = 1234567; // number to divide, and remainder at the end
uint32_t y = 0; // result
int bit = 31; // current bit
printf("X=%u X/3=%u\n",x,x/3); // the '/3' is for testing
while (bit>0)
{
printf("BIT=%d X=%u Y=%u\n",bit,x,y);
// decrement bit
int h = 1; while (1) { bit ^= h; if ( bit&h ) h <<= 1; else break; }
uint32_t r = x>>bit; // current remainder in 0..5
x ^= r<<bit; // remove R bits from X
if (r >= 3) y |= 1<<bit; // new output bit
x |= mod3[r]<<bit; // new remainder inserted in X
}
printf("Y=%u\n",y);
}
通过使用eval和字符串连接来在“幕后”使用/操作符是欺骗吗?
例如,在Javacript中,您可以这样做
function div3 (n) {
var div = String.fromCharCode(47);
return eval([n, div, 3].join(""));
}
很有趣的是,没有人回答一个泛泛的划分:
/* For the given integer find the position of MSB */
int find_msb_loc(unsigned int n)
{
if (n == 0)
return 0;
int loc = sizeof(n) * 8 - 1;
while (!(n & (1 << loc)))
loc--;
return loc;
}
/* Assume both a and b to be positive, return a/b */
int divide_bitwise(const unsigned int a, const unsigned int b)
{
int int_size = sizeof(unsigned int) * 8;
int b_msb_loc = find_msb_loc(b);
int d = 0; // dividend
int r = 0; // reminder
int t_a = a;
int t_a_msb_loc = find_msb_loc(t_a);
int t_b = b << (t_a_msb_loc - b_msb_loc);
int i;
for(i = t_a_msb_loc; i >= b_msb_loc; i--) {
if (t_a > t_b) {
d = (d << 1) | 0x1;
t_a -= t_b; // Not a bitwise operatiion
t_b = t_b >> 1;
}
else if (t_a == t_b) {
d = (d << 1) | 0x1;
t_a = 0;
}
else { // t_a < t_b
d = d << 1;
t_b = t_b >> 1;
}
}
r = t_a;
printf("==> %d %d\n", d, r);
return d;
}
按位加法已经在其中一个答案中给出,所以跳过它。
要将一个32位数字除以3,可以将其乘以0x55555556,然后取64位结果的前32位。
现在剩下要做的就是使用位运算和移位来实现乘法…
这是Python中的,基本上,字符串比较和一个状态机。
def divide_by_3(input):
to_do = {}
enque_index = 0
zero_to_9 = (0, 1, 2, 3, 4, 5, 6, 7, 8, 9)
leave_over = 0
for left_over in (0, 1, 2):
for digit in zero_to_9:
# left_over, digit => enque, leave_over
to_do[(left_over, digit)] = (zero_to_9[enque_index], leave_over)
if leave_over == 0:
leave_over = 1
elif leave_over == 1:
leave_over = 2
elif leave_over == 2 and enque_index != 9:
leave_over = 0
enque_index = (1, 2, 3, 4, 5, 6, 7, 8, 9)[enque_index]
answer_q = []
left_over = 0
digits = list(str(input))
if digits[0] == "-":
answer_q.append("-")
digits = digits[1:]
for digit in digits:
enque, left_over = to_do[(left_over, int(digit))]
if enque or len(answer_q):
answer_q.append(enque)
answer = 0
if len(answer_q):
answer = int("".join([str(a) for a in answer_q]))
return answer