我需要对一个整数a数组q(最多100,000个)进行异或.即如果我正在循环,我会
异或q [0]
异或[1]
.....
异或q [100000]
(100,000次)
我将有一系列这样的aXORed.
我正在编写一个控制台应用程序,它将传递所需的输入.
我正在使用内置的C#^操作符来执行XOR操作.还有其他方法吗?
将整数转换为字节数组然后对每个位进行异或并找出最终结果是一个好主意吗?
输入(不要保持两行之间的空格)
1
15 8
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
10 6 10
1023 7 7
33 5 8
182 5 10
181 1 13
5 10 15
99 8 9
33 10 14
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
namespace XOR
{
class Solution
{
static void Main(string[] args)
{
List<TestCase> testCases = ReadLine();
//Console.WriteLine(DateTime.Now.ToLongTimeString());
CalculationManager calculationManager = new CalculationManager();
foreach (var testCase in testCases)
{
var ints = testCase.Queries.AsParallel().Select(query => calculationManager.Calculate(query, testCase.SequenceOfIntegers)).ToList();
ints.ForEach(Console.WriteLine);
}
//Console.WriteLine(DateTime.Now.ToLongTimeString());
//Console.ReadLine();
}
private static List<TestCase> ReadLine()
{
int noOfTestCases = Convert.ToInt32(Console.ReadLine());
var testCases = new List<TestCase>();
for (int i = 0; i < noOfTestCases; i++)
{
string firstLine = Console.ReadLine();
string[] firstLineSplit = firstLine.Split(' ');
int N = Convert.ToInt32(firstLineSplit[0]);
int Q = Convert.ToInt32(firstLineSplit[1]);
var testCase = new TestCase
{
Queries = new List<Query>(),
SequenceOfIntegers = ReadLineAndGetSequenceOfIntegers()
};
for (int j = 0; j < Q; j++)
{
var buildQuery = ReadLineAndBuildQuery();
testCase.Queries.Add(buildQuery);
}
testCases.Add(testCase);
}
return testCases;
}
private static List<int> ReadLineAndGetSequenceOfIntegers()
{
string secondLine = Console.ReadLine();
List<int> sequenceOfIntegers = secondLine.Split(' ').ToArray().Select(x => Convert.ToInt32(x)).ToList();
return sequenceOfIntegers;
}
private static Query ReadLineAndBuildQuery()
{
var query = Console.ReadLine();
List<int> queryIntegers = query.Split(' ').ToArray().Select(x => Convert.ToInt32(x)).ToList();
Query buildQuery = ReadLineAndBuildQuery(queryIntegers[0], queryIntegers[1], queryIntegers[2]);
return buildQuery;
}
private static Query ReadLineAndBuildQuery(int a, int p, int q)
{
return new Query { a = a, p = p, q = q };
}
}
class CalculationManager
{
public int Calculate(Query query, List<int> sequenceOfIntegers)
{
var possibleIntegersToCalculate = FindPossibleIntegersToCalculate(sequenceOfIntegers, query.p, query.q);
int maxXorValue = possibleIntegersToCalculate.AsParallel().Max(x => x ^ query.a);
return maxXorValue;
}
private IEnumerable<int> FindPossibleIntegersToCalculate(List<int> sequenceOfIntegers, int p, int q)
{
return sequenceOfIntegers.GetRange(p - 1, (q - p) + 1).Distinct().ToArray();
}
}
class TestCase
{
public List<int> SequenceOfIntegers { get; set; }
public List<Query> Queries { get; set; }
}
class Query
{
public int a { get; set; }
public int p { get; set; }
public int q { get; set; }
}
}
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Sim*_*Var 16
使用^逐位xor运算符是xor整数的最快方法.
该操作被转换为单个原子处理器操作.
正如你在反汇编中看到的那样:
int i = 4;
00000029 mov dword ptr [ebp-3Ch],4
i ^= 3;
00000030 xor dword ptr [ebp-3Ch],3
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因此,如果您希望使代码运行得更快,则应更改算法/方法(如Marc Gravell所建议),而不是xor方法.
我甚至尝试过的唯一的事情(如果有理由认为int方法太慢)将使用unsafe代码将每个处理int[]为a long*,并使用64位算法(再次使用^)而不是32,一半迭代,少一点间接.IIRC就像我对某些Web套接字代码所做的那样(为客户端到服务器消息应用Web套接字掩码是一个批量异或).显然,你需要小心最后几个字节.