如何:使用并行容器提高效率

本主题说明如何使用并行容器以并行方式高效地存储和访问数据。

代码示例以并行方式计算一组质数和 Carmichael 数。然后,对于每个 Carmichael 数,此代码将计算该数的质数因子。

示例

下面的示例显示了 is_prime 函数和 is_carmichael 函数,前者可确定输入值是否为质数,后者可确定输入值是否为 Carmichael 数。

// Determines whether the input value is prime.
bool is_prime(int n)
{
   if (n < 2)
      return false;
   for (int i = 2; i < n; ++i)
   {
      if ((n % i) == 0)
         return false;
   }
   return true;
}

// Determines whether the input value is a Carmichael number.
bool is_carmichael(const int n) 
{
   if (n < 2) 
      return false;

   int k = n;
   for (int i = 2; i <= k / i; ++i) 
   {
      if (k % i == 0) 
      {
         if ((k / i) % i == 0) 
            return false;
         if ((n - 1) % (i - 1) != 0)
            return false;
         k /= i;
         i = 1;
      }
   }
   return k != n && (n - 1) % (k - 1) == 0;
}

下面的示例使用 is_primeis_carmichael 函数计算多组质数和 Carmichael 数。该示例使用 concurrency::parallel_invokeconcurrency::parallel_for 算法来计算每个并行的设置。有关并行算法的更多信息,请参见并行算法

本示例使用 concurrency::concurrent_queue 对象来保存 Carmichael 的一组数字,因为它以后将使用该对象为工作队列。它使用 concurrency::concurrent_vector 对象来保存质数的集,因为它以后将循环访问此设置为查找主要因素。

// The maximum number to test.
const int max = 10000000;

// Holds the Carmichael numbers that are in the range [0, max).
concurrent_queue<int> carmichaels;

// Holds the prime numbers that are in the range  [0, sqrt(max)).
concurrent_vector<int> primes;

// Generate the set of Carmichael numbers and the set of prime numbers
// in parallel.
parallel_invoke(
   [&] {
      parallel_for(0, max, [&](int i) {
         if (is_carmichael(i)) {
            carmichaels.push(i);
         }
      });
   },
   [&] {
      parallel_for(0, int(sqrt(static_cast<double>(max))), [&](int i) {
         if (is_prime(i)) {
            primes.push_back(i);
         }
      });
   });

下面的示例演示 prime_factors_of 函数,该函数使用试除法来查找给定值的所有质数因子。

此函数使用 concurrency::parallel_for_each 算法来循环访问集合的质数。concurrent_vector 对象使并行循环能够同时向结果中添加多个质数因子。

// Finds all prime factors of the given value.
concurrent_vector<int> prime_factors_of(int n, 
   const concurrent_vector<int>& primes)
{
   // Holds the prime factors of n.
   concurrent_vector<int> prime_factors;

   // Use trial division to find the prime factors of n.
   // Every prime number that divides evenly into n is a prime factor of n.
   const int max = sqrt(static_cast<double>(n));
   parallel_for_each(begin(primes), end(primes), [&](int prime)
   {
      if (prime <= max)
      {         
         if ((n % prime) == 0)
            prime_factors.push_back(prime);
      }
   });

   return prime_factors;
}

此示例通过调用 prime_factors_of 函数计算 Carmichael 数的质数因子来处理 Carmichael 数队列中的每个元素。此示例使用任务组以并行方式执行此任务。有关任务组的更多信息,请参见任务并行(并发运行时)

此示例将输出每个具有四个以上的质数因子的 Carmichael 数的质数因子。

// Use a task group to compute the prime factors of each 
// Carmichael number in parallel.
task_group tasks;

int carmichael;
while (carmichaels.try_pop(carmichael))
{
   tasks.run([carmichael,&primes] 
   {
      // Compute the prime factors.
      auto prime_factors = prime_factors_of(carmichael, primes);

      // For brevity, print the prime factors for the current number only
      // if there are more than 4.
      if (prime_factors.size() > 4)
      {
         // Sort and then print the prime factors.
         sort(begin(prime_factors), end(prime_factors));

         wstringstream ss;
         ss << L"Prime factors of " << carmichael << L" are:";

         for_each (begin(prime_factors), end(prime_factors), 
            [&](int prime_factor) { ss << L' ' << prime_factor; });
         ss << L'.' << endl;

         wcout << ss.str();
      }
   });
}

// Wait for the task group to finish.
tasks.wait();

下面的代码演示完整示例,该示例使用并行容器计算 Carmichael 数的质数因子。

// carmichael-primes.cpp
// compile with: /EHsc
#include <ppl.h>
#include <concurrent_queue.h>
#include <concurrent_vector.h>
#include <iostream>
#include <sstream>

using namespace concurrency;
using namespace std;

// Determines whether the input value is prime.
bool is_prime(int n)
{
   if (n < 2)
      return false;
   for (int i = 2; i < n; ++i)
   {
      if ((n % i) == 0)
         return false;
   }
   return true;
}

// Determines whether the input value is a Carmichael number.
bool is_carmichael(const int n) 
{
   if (n < 2) 
      return false;

   int k = n;
   for (int i = 2; i <= k / i; ++i) 
   {
      if (k % i == 0) 
      {
         if ((k / i) % i == 0) 
            return false;
         if ((n - 1) % (i - 1) != 0)
            return false;
         k /= i;
         i = 1;
      }
   }
   return k != n && (n - 1) % (k - 1) == 0;
}

// Finds all prime factors of the given value.
concurrent_vector<int> prime_factors_of(int n, 
   const concurrent_vector<int>& primes)
{
   // Holds the prime factors of n.
   concurrent_vector<int> prime_factors;

   // Use trial division to find the prime factors of n.
   // Every prime number that divides evenly into n is a prime factor of n.
   const int max = sqrt(static_cast<double>(n));
   parallel_for_each(begin(primes), end(primes), [&](int prime)
   {
      if (prime <= max)
      {         
         if ((n % prime) == 0)
            prime_factors.push_back(prime);
      }
   });

   return prime_factors;
}

int wmain()
{
   // The maximum number to test.
   const int max = 10000000;

   // Holds the Carmichael numbers that are in the range [0, max).
   concurrent_queue<int> carmichaels;

   // Holds the prime numbers that are in the range  [0, sqrt(max)).
   concurrent_vector<int> primes;

   // Generate the set of Carmichael numbers and the set of prime numbers
   // in parallel.
   parallel_invoke(
      [&] {
         parallel_for(0, max, [&](int i) {
            if (is_carmichael(i)) {
               carmichaels.push(i);
            }
         });
      },
      [&] {
         parallel_for(0, int(sqrt(static_cast<double>(max))), [&](int i) {
            if (is_prime(i)) {
               primes.push_back(i);
            }
         });
      });

   // Use a task group to compute the prime factors of each 
   // Carmichael number in parallel.
   task_group tasks;

   int carmichael;
   while (carmichaels.try_pop(carmichael))
   {
      tasks.run([carmichael,&primes] 
      {
         // Compute the prime factors.
         auto prime_factors = prime_factors_of(carmichael, primes);

         // For brevity, print the prime factors for the current number only
         // if there are more than 4.
         if (prime_factors.size() > 4)
         {
            // Sort and then print the prime factors.
            sort(begin(prime_factors), end(prime_factors));

            wstringstream ss;
            ss << L"Prime factors of " << carmichael << L" are:";

            for_each (begin(prime_factors), end(prime_factors), 
               [&](int prime_factor) { ss << L' ' << prime_factor; });
            ss << L'.' << endl;

            wcout << ss.str();
         }
      });
   }

   // Wait for the task group to finish.
   tasks.wait();
}

此示例产生下面的示例输出。

Prime factors of 9890881 are: 7 11 13 41 241.
Prime factors of 825265 are: 5 7 17 19 73.
Prime factors of 1050985 are: 5 13 19 23 37.

编译代码

将示例代码复制并将其粘贴在 Visual Studio 项目中,或将它粘贴到一个文件,名为 carmichael primes.cpp ,然后在 Visual Studio 命令提示符窗口中运行以下命令。

cl.exe /EHsc carmichael-primes.cpp

请参见

参考

concurrent_vector 类

concurrent_queue 类

parallel_invoke 函数

parallel_for 函数

task_group 类

概念

并行容器和对象

任务并行(并发运行时)