如何将一个数除3而不使用*、/、+、-、%等运算符?
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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
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。
其他回答
第一:
x/3 = (x/4) / (1-1/4)
然后求x/(1 - y)
x/(1-1/y)
= x * (1+y) / (1-y^2)
= x * (1+y) * (1+y^2) / (1-y^4)
= ...
= x * (1+y) * (1+y^2) * (1+y^4) * ... * (1+y^(2^i)) / (1-y^(2^(i+i))
= x * (1+y) * (1+y^2) * (1+y^4) * ... * (1+y^(2^i))
y = 1/4:
int div3(int x) {
x <<= 6; // need more precise
x += x>>2; // x = x * (1+(1/2)^2)
x += x>>4; // x = x * (1+(1/2)^4)
x += x>>8; // x = x * (1+(1/2)^8)
x += x>>16; // x = x * (1+(1/2)^16)
return (x+1)>>8; // as (1-(1/2)^32) very near 1,
// we plus 1 instead of div (1-(1/2)^32)
}
虽然它使用了+,但有人已经实现了按位操作的add。
没有反复检查这个答案是否已经发表。如果程序需要扩展到浮点数,可以将这些数字乘以所需精度的10*数,然后可以再次应用下面的代码。
#include <stdio.h>
int main()
{
int aNumber = 500;
int gResult = 0;
int aLoop = 0;
int i = 0;
for(i = 0; i < aNumber; i++)
{
if(aLoop == 3)
{
gResult++;
aLoop = 0;
}
aLoop++;
}
printf("Reulst of %d / 3 = %d", aNumber, gResult);
return 0;
}
如果你提醒自己标准的学校除法方法,用二进制来做,你会发现在3的情况下,你只是在有限的一组值中除法和减法(在这种情况下,从0到5)。这些可以用switch语句处理,以摆脱算术运算符。
static unsigned lamediv3(unsigned n)
{
unsigned result = 0, remainder = 0, mask = 0x80000000;
// Go through all bits of n from MSB to LSB.
for (int i = 0; i < 32; i++, mask >>= 1)
{
result <<= 1;
// Shift in the next bit of n into remainder.
remainder = remainder << 1 | !!(n & mask);
// Divide remainder by 3, update result and remainer.
// If remainder is less than 3, it remains intact.
switch (remainder)
{
case 3:
result |= 1;
remainder = 0;
break;
case 4:
result |= 1;
remainder = 1;
break;
case 5:
result |= 1;
remainder = 2;
break;
}
}
return result;
}
#include <cstdio>
int main()
{
// Verify for all possible values of a 32-bit unsigned integer.
unsigned i = 0;
do
{
unsigned d = lamediv3(i);
if (i / 3 != d)
{
printf("failed for %u: %u != %u\n", i, d, i / 3);
return 1;
}
}
while (++i != 0);
}
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。
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