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
int div3(int x)
{
int reminder = abs(x);
int result = 0;
while(reminder >= 3)
{
result++;
reminder--;
reminder--;
reminder--;
}
return result;
}
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。
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
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。
使用itoa转换为以3为基数的字符串。去掉最后一个小调,转换回10进制。
// Note: itoa is non-standard but actual implementations
// don't seem to handle negative when base != 10.
int div3(int i) {
char str[42];
sprintf(str, "%d", INT_MIN); // Put minus sign at str[0]
if (i>0) // Remove sign if positive
str[0] = ' ';
itoa(abs(i), &str[1], 3); // Put ternary absolute value starting at str[1]
str[strlen(&str[1])] = '\0'; // Drop last digit
return strtol(str, NULL, 3); // Read back result
}
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。