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
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
其他回答
#include <stdio.h>
typedef struct { char a,b,c; } Triple;
unsigned long div3(Triple *v, char *r) {
if ((long)v <= 2)
return (unsigned long)r;
return div3(&v[-1], &r[1]);
}
int main() {
unsigned long v = 21;
int r = div3((Triple*)v, 0);
printf("%ld / 3 = %d\n", v, r);
return 0;
}
很好bc:
$ num=1337; printf "scale=5;${num}\x2F3;\n" | bc
445.66666
首先是我想到的。
irb(main):101:0> div3 = -> n { s = '%0' + n.to_s + 's'; (s % '').gsub(' ', ' ').size }
=> #<Proc:0x0000000205ae90@(irb):101 (lambda)>
irb(main):102:0> div3[12]
=> 4
irb(main):103:0> div3[666]
=> 222
编辑:对不起,我没有注意到标签c。但是你可以使用字符串格式的想法,我猜…
你可以考虑用图或树状结构来解决这个问题。基本上生成的顶点数与要除以3的数一样多。然后继续将每个未配对的顶点与其他两个顶点配对。
粗糙的伪代码:
function divide(int num)
while(num!=0)
Add a new vertice to vertiexList.
num--
quotient = 0
for each in vertexList(lets call this vertex A)
if vertexList not empty
Add an edge between A and another vertex(say B)
else
your Remainder is 1 and Quotient is quotient
if vertexList not empty
Add an edge between A and another vertex(say C)
else
your remainder is 2 and Quotient is quotient
quotient++
remove A, B, C from vertexList
Remainder is 0 and Quotient is quotient
这显然是可以优化的,复杂度取决于你的数字有多大,但它应该工作,只要你能做++和——。 这就像数更酷的东西一样。
使用cblas,作为OS X加速框架的一部分。
[02:31:59] [william@relativity ~]$ cat div3.c
#import <stdio.h>
#import <Accelerate/Accelerate.h>
int main() {
float multiplicand = 123456.0;
float multiplier = 0.333333;
printf("%f * %f == ", multiplicand, multiplier);
cblas_sscal(1, multiplier, &multiplicand, 1);
printf("%f\n", multiplicand);
}
[02:32:07] [william@relativity ~]$ clang div3.c -framework Accelerate -o div3 && ./div3
123456.000000 * 0.333333 == 41151.957031