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
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。
其他回答
这是我小时候爷爷教我的一个方法。它需要+和/运算符,但计算起来很简单。
把每个数字相加,然后看看它是否是3的倍数。
但这种方法适用于大于12的数字。
例如:36岁,
3+6=9,是3的倍数。
42,
4+2=6,是3的倍数。
使用黑客的喜悦魔术数字计算器
int divideByThree(int num)
{
return (fma(num, 1431655766, 0) >> 32);
}
其中fma是在math.h头文件中定义的标准库函数。
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。
在PHP中使用BC数学:
<?php
$a = 12345;
$b = bcdiv($a, 3);
?>
MySQL(来自Oracle的采访)
> SELECT 12345 DIV 3;
帕斯卡:
a:= 12345;
b:= a div 3;
X86-64汇编语言:
mov r8, 3
xor rdx, rdx
mov rax, 12345
idiv r8
这是一个执行所需操作的简单函数。但是它需要+操作符,所以你所要做的就是用位操作符来加值:
// replaces the + operator
int add(int x, int y)
{
while (x) {
int t = (x & y) << 1;
y ^= x;
x = t;
}
return y;
}
int divideby3(int num)
{
int sum = 0;
while (num > 3) {
sum = add(num >> 2, sum);
num = add(num >> 2, num & 3);
}
if (num == 3)
sum = add(sum, 1);
return sum;
}
正如吉姆评论的那样,这是可行的,因为:
N = 4 * a + b N / 3 = a + (a + b) / 3 sum += an = a + b,然后迭代 当a == 0 (n < 4)时,sum += floor(n / 3);即1,如果n == 3,否则为0