任意基数 b 的数的补码
在这篇文章中，讨论了一种寻找任意基数补码的一般方法
。
寻找(b-1)补语的步骤:寻找(b-1)补语，
- 用基数为
![b]( "Rendered by QuickLaTeX.com")
的数字系统中最大的数字减去数字的每个数字。 - 例如，如果数字是以 9 为基数的三位数，则从 888 中减去该数字，因为 8 是以 9 为基数的数字系统中最大的数字。
- 得到的结果是(b-1)的(8 的补码)补码。
的数字系统中最大的数字减去数字的每个数字。求 b 的补码的步骤:要求 b 的补码，只需在计算出的(b-1)的补码上加 1 即可。 现在，这适用于任何存在于数系中的基数。它可以用熟悉的 1 和 2 补码来测试。 示例 :
Let the number be 10111 base 2 (b=2)
Then, 1's complement will be 01000 (b-1)
2's complement will be 01001 (b)
Taking a number with Octal base:
Let the number be -456.
Then 7's complement will be 321
and 8's complement will be 322
以下是上述思路的实现:
C++
// CPP program to find complement of a
// number with any base b
#include<iostream>
#include<cmath>
using namespace std;
// Function to find (b-1)'s complement
int prevComplement(int n, int b)
{
int maxDigit, maxNum = 0, digits = 0, num = n;
// Calculate number of digits
// in the given number
while(n!=0)
{
digits++;
n = n/10;
}
// Largest digit in the number
// system with base b
maxDigit = b-1;
// Largest number in the number
// system with base b
while(digits--)
{
maxNum = maxNum*10 + maxDigit;
}
// return Complement
return maxNum - num;
}
// Function to find b's complement
int complement(int n, int b)
{
// b's complement = (b-1)'s complement + 1
return prevComplement(n,b) + 1;
}
// Driver code
int main()
{
cout << prevComplement(25, 7)<<endl;
cout << complement(25, 7);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program to find complement
// of a number with any base b
class GFG
{
// Function to find (b-1)'s complement
static int prevComplement(int n, int b)
{
int maxDigit, maxNum = 0,
digits = 0, num = n;
// Calculate number of digits
// in the given number
while(n != 0)
{
digits++;
n = n / 10;
}
// Largest digit in the number
// system with base b
maxDigit = b - 1;
// Largest number in the number
// system with base b
while((digits--) > 0)
{
maxNum = maxNum * 10 + maxDigit;
}
// return Complement
return maxNum - num;
}
// Function to find b's complement
static int complement(int n, int b)
{
// b's complement = (b-1)'s
// complement + 1
return prevComplement(n, b) + 1;
}
// Driver code
public static void main(String args[])
{
System.out.println(prevComplement(25, 7));
System.out.println(complement(25, 7));
}
}
// This code is contributed
// by Kirti_Mangal
Python 3
# Python 3 program to find
# complement of a number
# with any base b
# Function to find
# (b-1)'s complement
def prevComplement(n, b) :
maxNum, digits, num = 0, 0, n
# Calculate number of digits
# in the given number
while n > 1 :
digits += 1
n = n // 10
# Largest digit in the number
# system with base b
maxDigit = b - 1
# Largest number in the number
# system with base b
while digits :
maxNum = maxNum * 10 + maxDigit
digits -= 1
# return Complement
return maxNum - num
# Function to find b's complement
def complement(n, b) :
# b's complement = (b-1)'s
# complement + 1
return prevComplement(n, b) + 1
# Driver code
if __name__ == "__main__" :
# Function calling
print(prevComplement(25, 7))
print(complement(25, 7))
# This code is contributed
# by ANKITRAI1
C
// C# program to find complement
// of a number with any base b
class GFG
{
// Function to find (b-1)'s complement
static int prevComplement(int n, int b)
{
int maxDigit, maxNum = 0,
digits = 0, num = n;
// Calculate number of digits
// in the given number
while(n != 0)
{
digits++;
n = n / 10;
}
// Largest digit in the number
// system with base b
maxDigit = b - 1;
// Largest number in the number
// system with base b
while((digits--) > 0)
{
maxNum = maxNum * 10 + maxDigit;
}
// return Complement
return maxNum - num;
}
// Function to find b's complement
static int complement(int n, int b)
{
// b's complement = (b-1)'s
// complement + 1
return prevComplement(n, b) + 1;
}
// Driver code
public static void Main()
{
System.Console.WriteLine(prevComplement(25, 7));
System.Console.WriteLine(complement(25, 7));
}
}
// This code is contributed
// by mits
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP program to find complement
// of a number with any base b
// Function to find (b-1)'s complement
function prevComplement($n, $b)
{
$maxNum = 0;
$digits = 0;
$num = $n;
// Calculate number of digits
// in the given number
while((int)$n != 0)
{
$digits++;
$n = $n / 10;
}
// Largest digit in the number
// system with base b
$maxDigit = $b - 1;
// Largest number in the number
// system with base b
while($digits--)
{
$maxNum = $maxNum * 10 +
$maxDigit;
}
// return Complement
return $maxNum - $num;
}
// Function to find b's complement
function complement($n, $b)
{
// b's complement = (b-1)'s
// complement + 1
return prevComplement($n, $b) + 1;
}
// Driver code
echo prevComplement(25, 7), "\n";
echo(complement(25, 7));
// This code is contributed
// by Smitha
?>
java 描述语言
<script>
// Javascript program to find complement
// of a number with any base b
// Function to find (b-1)'s complement
function prevComplement(n , b) {
var maxDigit, maxNum = 0, digits = 0,
num = n;
// Calculate number of digits
// in the given number
while (n != 0) {
digits++;
n = parseInt(n / 10);
}
// Largest digit in the number
// system with base b
maxDigit = b - 1;
// Largest number in the number
// system with base b
while ((digits--) > 0) {
maxNum = maxNum * 10 + maxDigit;
}
// return Complement
return maxNum - num;
}
// Function to find b's complement
function complement(n , b) {
// b's complement = (b-1)'s
// complement + 1
return prevComplement(n, b) + 1;
}
// Driver code
document.write(prevComplement(25, 7)+"<br/>");
document.write(complement(25, 7));
// This code is contributed by todaysgaurav
</script>
Output:
41
42
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