HOW TO：使用平行容器提高效率

發行項
08/12/2011

本主題將示範如何使用平行容器，以平行方式有效率地儲存和存取資料。

此範例程式碼會以平行方式計算質數和 Carmichael 數字的集合。然後，此程式碼會針對每個 Carmichael 數字，計算該數字的質因數。

範例

下列範例顯示了 is_prime 函式 (它會判斷輸入值是否為質數) 以及 is_carmichael 函式 (它會判斷輸入值是否為 Carmichael 數字)。

// 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;
}

下列範例會使用 is_prime 和 is_carmichael 函式來計算質數和 Carmichael 數字的集合。此範例會使用 Concurrency::parallel_invoke 和 Concurrency::parallel_for 演算法，以平行方式計算每個集合。如需平行演算法的詳細資訊，請參閱平行演算法。

這個範例會使用 Concurrency::concurrent_queue 物件來保存 Carmichael 數字的集合，因為它之後將使用該物件當做工作佇列。它會使用 Concurrency::concurrent_vector 物件來保存質數的集合，因為它之後將逐一查看此集合來尋找質因數。

// 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);
         }
      });
   });

下列範例顯示了 prime_factors_of 函式，它會使用試除法來尋找給定值的所有質因數。

此函式會使用 Concurrency::parallel_for_each 演算法來逐一查看質數的集合。 concurrent_vector 物件可讓平行迴圈以並行方式將質因數加入至結果。

// 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(primes.begin(), primes.end(), [&](int prime)
   {
      if (prime <= max)
      {         
         if ((n % prime) == 0)
            prime_factors.push_back(prime);
      }
   });

   return prime_factors;
}

這個範例會透過呼叫 prime_factors_of 函式來計算質因數，處理 Carmichael 數字佇列中的每個項目。它會使用工作群組，以平行方式執行這項工作。如需工作群組的詳細資訊，請參閱工作平行處理原則 (並行執行階段)。

這個範例會列印每個 Carmichael 數字的質因數 (如果該數字的質因數超過四個的話)。

// 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(prime_factors.begin(), prime_factors.end());

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

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

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

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

下列程式碼顯示了完整的範例，其中使用平行容器來計算 Carmichael 數字的質因數。

// 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(primes.begin(), primes.end(), [&](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(prime_factors.begin(), prime_factors.end());

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

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

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

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

這個範例 (Example) 產生下列範例 (Sample) 輸出。

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.

編譯程式碼

請複製範例程式碼，並將它貼在 Visual Studio 專案中，或貼在名為 carmichael-primes.cpp 的檔案中，然後在 Visual Studio 2010 的 [命令提示字元] 視窗中執行下列命令。

cl.exe /EHsc carmichael-primes.cpp

請參閱

共用方式為

HOW TO：使用平行容器提高效率

範例

編譯程式碼

請參閱

參考

概念

其他資源

其他資源