浙江大学PAT---Basic Level---1007----素数对猜想

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Diagnostics;

namespace PAT_1007
{
class Program
{
static void Main(string[] args)
{
//输入数据
//
Console.WriteLine("请输入数据:");
int input,i,count=0;
int[] prime1;
bool[] prime2;

//计算小于输入的素数
//
Stopwatch sw = new Stopwatch();
sw.Start();

input = Int32.Parse(Console.ReadLine());
prime1 = Prime1(input);
sw.Stop();
Console.WriteLine("方法一耗时：{0}", sw.ElapsedTicks);

sw.Restart();
prime2 = Prime2(input);
sw.Stop();
Console.WriteLine("方法二耗时：{0}", sw.ElapsedTicks);

//输出素数
//
i = 0;
Console.WriteLine("素数：");
while (prime1[i] != 0)
{
Console.Write("{0} ", prime1[i++]);
}
Console.Write("\n");

Console.WriteLine("素数：");
for (i = 0; i <= input; i++)
{
if (prime2[i])
Console.Write("{0} ", i);
}
Console.Write("\n");

//计算素数对
//
count = 0;
i = 1;
while (prime1[i] != 0)
{
if ((prime1[i] - prime1[i - 1]) == 2)
count++;
i++;
}
Console.Write("方法一素数对:");
Console.WriteLine(count);

count = 0;
for (i = 2; i <= input; i++)
{
if (prime2[i] && prime2[i - 2])
count++;
}
Console.Write("方法二素数对:");
Console.WriteLine(count);

Console.ReadKey();
}

/// <summary>
/// 最一般素数求解方法，理论：一个合数至少有一个小于sqrt（n）的因子
/// </summary>
/// <param name="n"></param>
/// <returns></returns>
private static int[] Prime1(int n)
{
int[] prime = new int
;
int i = 0, j=0,count = 0;

for (i = 2; i <= n; i++)
{
for (j = 2; j <= Math.Sqrt(i); j++)
if (i % j == 0) break;
if (j > Math.Sqrt(i)) prime[count++] = i;  //这个prime[]是int型，跟下面讲的不同。
}
return prime;
}

/// <summary>
/// 筛选法计算素数，理论：素数的倍数是合数
/// </summary>
/// <param name="n"></param>
/// <returns></returns>
private static bool[] Prime2(int n)
{
bool[] prime = new bool[n + 1];

for (int i = 0, j = 1; j <= n; i += 2, j += 2)
{
prime[i] = false;    //偶数
prime[j] = true;     //奇数
}

for (int i = 3; i <= Math.Sqrt(n); i+=2)
{
if (prime[i])
{
for (int j = i+i; j <= n; j += i)
{
prime[j] = false;
}
}
}
prime[1] = false;
prime[2] = true;

return prime;
}
}
}