如何解决如何在 C# 中执行 2 个非常大超过 50 位二进制字符串值的加法
我一直在考虑添加二进制数，其中二进制数是字符串形式，我们将这两个二进制数相加得到一个结果二进制数（以字符串形式）。
到目前为止我一直在使用这个：
long first = Convert.ToInt64(a,2);
long second = Convert.ToInt64(b,2);
long addresult = first + second;
string result = Convert.ToString(addresult,2);
return result;
由 Stackoverflow 提供：Binary addition of 2 values represented as strings
但是，现在我想添加远大于 long 数据类型范围的数字。
例如，十进制结果为 BigInteger 的二进制值，即如下所示的非常大的整数：
-
111101101000010111101000101010001010010010010110011010100001000010110110110000110001101等于149014059082024308625334669 -
1111001101011000001011000111100011101011110100101010010001110101011101010100101000001101000010000110001110100010011101011111111110110101100101110001010101011110001010000010111110011011等于23307765732196437339985049250988196614799400063289798555
至少我认为是这样。
由 Stackoverflow 提供：
我使用了以上链接中提供的或多或少完美的逻辑。
但是，有没有更紧凑的方法来添加给定的两个二进制字符串？
请告诉我，因为这个问题在我脑海中萦绕了一段时间。
解决方法
您可以利用之前使用的相同方案，但使用 BigInteger：
using System.Linq;
using System.Numerics;
...
BigInteger first = a.Aggregate(BigInteger.Zero,(s,item) => s * 2 + item - '0');
BigInteger second = b.Aggregate(BigInteger.Zero,item) => s * 2 + item - '0');
StringBuilder sb = new StringBuilder();
for (BigInteger addresult = first + second; addresult > 0; addresult /= 2)
sb.Append(addresult % 2);
if (sb.Length <= 0)
sb.Append('0');
string result = string.Concat(sb.ToString().Reverse());
这个问题很怀旧 - 谢谢。请注意，我的代码是解释性和低效的，几乎没有验证，但它适用于您的示例。你绝对不想在现实世界的解决方案中使用这样的东西，这只是为了说明原则上的二进制加法。
BinaryString#1
111101101000010111101000101010001010010010010110011010100001000010110110110000110001101
decimal:149014059082024308625334669
BinaryString#2
1111001101011000001011000111100011101011110100101010010001110101011101010100101000001101000010000110001110100010011101011111111110110101100101110001010101011110001010000010111110011011
decimal:23307765732196437339985049250988196614799400063289798555
Calculated Sum
1111001101011000001011000111100011101011110100101010010001110101011101010100101000001101000010001101111011100101011010100101010000000111111000100100101001100110100000111001000100101000
decimal:23307765732196437339985049251137210673881424371915133224
Check
23307765732196437339985049251137210673881424371915133224
decimal:23307765732196437339985049251137210673881424371915133224
这是代码
using System;
using System.Linq;
using System.Numerics;
namespace ConsoleApp3
{
class Program
{
// return 0 for '0' and 1 for '1' (C# chars promotion to ints)
static int CharAsInt(char c) { return c - '0'; }
// and vice-versa
static char IntAsChar(int bit) { return (char)('0' + bit); }
static string BinaryStringAdd(string x,string y)
{
// get rid of spaces
x = x.Trim();
y = y.Trim();
// check if valid binaries
if (x.Any(c => c != '0' && c != '1') || y.Any(c => c != '0' && c != '1'))
throw new ArgumentException("binary representation may contain only '0' and '1'");
// align on right-most bit
if (x.Length < y.Length)
x = x.PadLeft(y.Length,'0');
else
y = y.PadLeft(x.Length,'0');
// NNB: the result may require one more bit than the longer of the two input strings (carry at the end),let's not forget this
var result = new char[x.Length];
// add from least significant to most significant (right to left)
var i = result.Length;
var carry = '0';
while (--i >= 0)
{
// to add x[i],y[i] and carry
// - if 2 or 3 bits are set then we carry '1' again (otherwise '0')
// - if the number of set bits is odd the sum bit is '1' otherwise '0'
var countSetBits = CharAsInt(x[i]) + CharAsInt(y[i]) + CharAsInt(carry);
carry = countSetBits > 1 ? '1' : '0';
result[i] = countSetBits == 1 || countSetBits == 3 ? '1' : '0';
}
// now to make this byte[] a string
var ret = new string(result);
// remember that final carry?
return carry == '1' ? carry + ret : ret;
}
static BigInteger BigIntegerFromBinaryString(string s)
{
var biRet = new BigInteger(0);
foreach (var t in s)
{
biRet = biRet << 1;
if (t == '1')
biRet += 1;
}
return biRet;
}
static void Main(string[] args)
{
var s1 = "111101101000010111101000101010001010010010010110011010100001000010110110110000110001101";
var s2 = "1111001101011000001011000111100011101011110100101010010001110101011101010100101000001101000010000110001110100010011101011111111110110101100101110001010101011110001010000010111110011011";
var sum = BinaryStringAdd(s1,s2);
var bi1 = BigIntegerFromBinaryString(s1);
var bi2 = BigIntegerFromBinaryString(s2);
var bi3 = bi1 + bi2;
Console.WriteLine($"BinaryString#1\n {s1}\n decimal:{bi1}");
Console.WriteLine($"BinaryString#2\n {s2}\n decimal:{bi2}");
Console.WriteLine($"Calculated Sum\n {sum}\n decimal:{BigIntegerFromBinaryString(sum)}");
Console.WriteLine($"Check\n {bi3}\n decimal:{bi3}");
Console.ReadKey();
}
}
}
我将在 AlanK 旁边添加一个替代解决方案，作为一个示例，说明您如何在不将数字转换为某种形式的整数的情况下进行添加。
static string BinaryStringAdd(string b1,string b2)
{
char[] c = new char[Math.Max(b1.Length,b2.Length) + 1];
int carry = 0;
for (int i = 1; i <= c.Length; i++)
{
int d1 = i <= b1.Length ? b1[^i] : 48;
int d2 = i <= b2.Length ? b2[^i] : 48;
int sum = carry + (d1-48) + (d2-48);
if (sum == 3)
{
sum = 1;
carry = 1;
}
else if (sum == 2)
{
sum = 0;
carry = 1;
}
else
{
carry = 0;
}
c[^i] = (char) (sum+48);
}
return c[0] == '0' ? String.Join("",c)[1..] : String.Join("",c);
}
请注意，此解决方案比 Alan 的解决方案慢约 10%（至少对于此测试用例），并假设字符串到达格式正确的方法。
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