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
这是可行的:
smegma$ curl http://www.wolframalpha.com/input/?i=14+divided+by+3 2>/dev/null | gawk 'match($0, /link to /input/\?i=([0-9.+-]+)/, ary) { print substr( $0, ary[1, "start"], ary[1, "length"] )}' 4.6666666666666666666666666666666666666666666666666666
只要把你的数字换成“14”和“3”就行了。
其他回答
Yet another solution. This should handle all ints (including negative ints) except the min value of an int, which would need to be handled as a hard coded exception. This basically does division by subtraction but only using bit operators (shifts, xor, & and complement). For faster speed, it subtracts 3 * (decreasing powers of 2). In c#, it executes around 444 of these DivideBy3 calls per millisecond (2.2 seconds for 1,000,000 divides), so not horrendously slow, but no where near as fast as a simple x/3. By comparison, Coodey's nice solution is about 5 times faster than this one.
public static int DivideBy3(int a) {
bool negative = a < 0;
if (negative) a = Negate(a);
int result;
int sub = 3 << 29;
int threes = 1 << 29;
result = 0;
while (threes > 0) {
if (a >= sub) {
a = Add(a, Negate(sub));
result = Add(result, threes);
}
sub >>= 1;
threes >>= 1;
}
if (negative) result = Negate(result);
return result;
}
public static int Negate(int a) {
return Add(~a, 1);
}
public static int Add(int a, int b) {
int x = 0;
x = a ^ b;
while ((a & b) != 0) {
b = (a & b) << 1;
a = x;
x = a ^ b;
}
return x;
}
这是c#,因为这是我手边的东西,但与c的区别应该很小。
这真的很简单。
if (number == 0) return 0;
if (number == 1) return 0;
if (number == 2) return 0;
if (number == 3) return 1;
if (number == 4) return 1;
if (number == 5) return 1;
if (number == 6) return 2;
(当然,为了简洁起见,我省略了一些程序。)如果程序员厌倦了把这些都输入,我相信他或她可以编写一个单独的程序来为他生成这些。我碰巧认识一个能大大简化他工作的人。
为什么我们不直接用在大学里学过的定义呢?结果可能效率低,但很清楚,因为乘法只是递归的减法,减法是加法,那么加法可以通过递归的异或/和逻辑端口组合来执行。
#include <stdio.h>
int add(int a, int b){
int rc;
int carry;
rc = a ^ b;
carry = (a & b) << 1;
if (rc & carry)
return add(rc, carry);
else
return rc ^ carry;
}
int sub(int a, int b){
return add(a, add(~b, 1));
}
int div( int D, int Q )
{
/* lets do only positive and then
* add the sign at the end
* inversion needs to be performed only for +Q/-D or -Q/+D
*/
int result=0;
int sign=0;
if( D < 0 ) {
D=sub(0,D);
if( Q<0 )
Q=sub(0,Q);
else
sign=1;
} else {
if( Q<0 ) {
Q=sub(0,Q);
sign=1;
}
}
while(D>=Q) {
D = sub( D, Q );
result++;
}
/*
* Apply sign
*/
if( sign )
result = sub(0,result);
return result;
}
int main( int argc, char ** argv )
{
printf( "2 plus 3=%d\n", add(2,3) );
printf( "22 div 3=%d\n", div(22,3) );
printf( "-22 div 3=%d\n", div(-22,3) );
printf( "-22 div -3=%d\n", div(-22,-3) );
printf( "22 div 03=%d\n", div(22,-3) );
return 0;
}
有人说……首先让它工作。注意,该算法应该适用于负Q…
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