我正在努力完成作業 MOHIBPIZ - PIZZA ( https://www.spoj.com/problems/MOHIBPIZ/ )。第二天我已經坐在上面了，我已經嘗試了所有可以在互聯網上找到的方法。放棄之前的最后一次機會是問你們是否有時間我正在使用由 davidsekar 創建的 InputOutput 類（https://github.com/davidsekar/C-sharp-Programming-IO/blob/master/ConsoleInOut/InputOutput。 cs）但我仍然有時間“超出時間限制”。:(
我嘗試了兩個回圈，但該函式的方法對我來說似乎更理想。在此先感謝您提供的所有提示、建議和答案。
這是代碼（ideone 上的鏈接：https : //ideone.com/ ）：
using System;
using System.IO;
public class Test
{
public static void Main()
{
InputOutput reader = new InputOutput();
StreamWriter _output = new StreamWriter(Console.OpenStandardOutput());
int T = reader.ReadInt();
for (int i = 0; i < T; i )
{
_output.WriteLine(Recursion(reader.ReadInt()));
}
_output.Flush();
}
private static int Recursion(int x)
{
if(x <= 1)
{
return 2;
}
else
{
return Recursion(x - 1) x;
}
}
#region Input Output Helper
public class InputOutput : System.IDisposable
{
private System.IO.Stream _readStream, _writeStream;
private int _readIdx, _bytesRead, _writeIdx, _inBuffSize, _outBuffSize;
private readonly byte[] _inBuff, _outBuff;
private readonly bool _bThrowErrorOnEof;
public void SetBuffSize(int n)
{
_inBuffSize = _outBuffSize = n;
}
public InputOutput(bool throwEndOfInputsError = false)
{
_readStream = System.Console.OpenStandardInput();
_writeStream = System.Console.OpenStandardOutput();
_readIdx = _bytesRead = _writeIdx = 0;
_inBuffSize = _outBuffSize = 1 << 22;
_inBuff = new byte[_inBuffSize];
_outBuff = new byte[_outBuffSize];
_bThrowErrorOnEof = throwEndOfInputsError;
}
public void SetFilePath(string strPath)
{
strPath = System.IO.Path.GetFullPath(strPath);
_readStream = System.IO.File.Open(strPath, System.IO.FileMode.Open);
}
public T ReadNumber<T>()
{
byte rb;
while ((rb = GetByte()) < '-')
;
var neg = false;
if (rb == '-')
{
neg = true;
rb = GetByte();
}
dynamic m = (T)Convert.ChangeType(rb - '0', typeof(T));
while (true)
{
rb = GetByte();
if (rb < '0')
break;
m = m * 10 (rb - '0');
}
return neg ? -m : m;
}
public int ReadInt()
{
byte readByte;
while ((readByte = GetByte()) < '-')
;
var neg = false;
if (readByte == '-')
{
neg = true;
readByte = GetByte();
}
var m = readByte - '0';
while (true)
{
readByte = GetByte();
if (readByte < '0')
break;
m = m * 10 (readByte - '0');
}
return neg ? -m : m;
}
public string ReadString()
{
return ReadString(' ');
}
public string ReadString(string delimiter)
{
return ReadString(delimiter[0]);
}
public string ReadString(char delimiter)
{
byte readByte;
while ((readByte = GetByte()) <= delimiter)
;
System.Text.StringBuilder sb = new System.Text.StringBuilder();
do
{
sb.Append((char)readByte);
} while ((readByte = GetByte()) > delimiter);
return sb.ToString();
}
[System.Runtime.CompilerServices.MethodImpl(System.Runtime.CompilerServices.MethodImplOptions.AggressiveInlining)]
private byte GetByte()
{
if (_readIdx >= _bytesRead)
{
_readIdx = 0;
_bytesRead = _readStream.Read(_inBuff, 0, _inBuffSize);
if (_bytesRead >= 1)
return _inBuff[_readIdx ];
if (_bThrowErrorOnEof)
throw new System.Exception("End Of Input");
_inBuff[_bytesRead ] = 0;
}
return _inBuff[_readIdx ];
}
public void WriteToBuffer(string s)
{
foreach (var b in System.Text.Encoding.ASCII.GetBytes(s))
{
if (_writeIdx == _outBuffSize)
Flush();
_outBuff[_writeIdx ] = b;
}
}
public void WriteLineToBuffer(string s)
{
WriteToBuffer(s);
if (_writeIdx == _outBuffSize)
Flush();
_outBuff[_writeIdx ] = 10;
}
public void WriteToBuffer(int c)
{
byte[] temp = new byte[10];
int tempidx = 0;
if (c < 0)
{
if (_writeIdx == _outBuffSize)
Flush();
_outBuff[_writeIdx ] = (byte)'-';
c = -c;
}
do
{
temp[tempidx ] = (byte)((c % 10) '0');
c /= 10;
} while (c > 0);
for (int i = tempidx - 1; i >= 0; i--)
{
if (_writeIdx == _outBuffSize)
Flush();
_outBuff[_writeIdx ] = temp[i];
}
}
public void WriteLineToBuffer(int c)
{
WriteToBuffer(c);
if (_writeIdx == _outBuffSize)
Flush();
_outBuff[_writeIdx ] = 10;
}
private void Flush()
{
_writeStream.Write(_outBuff, 0, _writeIdx);
_writeStream.Flush();
_writeIdx = 0;
}
public void Dispose()
{
Flush();
_writeStream.Close();
_readStream.Close();
}
}
#endregion Input Output Helper
}
uj5u.com熱心網友回復：
據我所知，你有一個眾所周知的Circle Division 問題；另見A000124序列：
n切割后的件數是(n * n n 2) / 2
這就是為什么我們可以把O(1)時間和空間復雜度
代碼：
private static int Solution(int n) => (int)(((long)n * n n 2) / 2);
在這里，我已經放置(long) n了 casen * n 超過了 int.MaxValue，何時(n * n n 2) / 2沒有。
編輯：我已經實作了int Solution(int n)基于當前代碼int Recursion(int x) 簽名的方法；但是如果有大的 測驗，n我們將有整數溢位。在這種情況下
private static long Solution(long n) =>
1 (n % 2 == 0 ? n / 2 * (n 1) : (n 1) / 2 * n);
在任意情況下，n我們必須使用BigInteger：
using System.Numerics;
...
private static BigInteger Solution(BigInteger n) =>
1 (n * n n) / 2;
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