如何：使用平行容器提高效率

發行項
11/22/2024

本主題說明如何使用平行容器，以有效率地平行儲存和存取數據。

此範例程式代碼會平行計算質數和 Carmichael 數位集。然後，針對每個 Carmichael 數位，程式代碼會計算該數位的主要因素。

範例：判斷輸入值是否為質數

下列範例顯示 is_prime 函式，這個函式會判斷輸入值是否為質數，而 is_carmichael 函式會判斷輸入值是否為 Carmichael 數位。

// 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;
}

範例：計算質數和 Carmichael 數位

下列範例會使用 is_prime 和 is_carmichael 函式來計算質數和 Carmichael 數位集。此範例會使用 concurrency：:p arallel_invoke 和 concurrency：:p arallel_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。

此範例會使用並行：：concurrent_queue 物件來保存一組 Carmichael 數字，因為它稍後會使用該對象作為工作佇列。它會使用並行：：concurrent_vector 物件來保存一組質數，因為它稍後會逐一查看此集合來尋找質數。

// 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);
         }
      });
   });

範例：尋找指定值的所有主要因素

下列範例顯示函 prime_factors_of 式，該函式會使用試用除法來尋找指定值的所有主要因素。

此函式會使用 concurrency：:p arallel_for_each 演算法逐一查看質數集合。物件 concurrent_vector 可讓平行迴圈同時將主要因素新增至結果。

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

   return prime_factors;
}

範例：處理 Carmichael 數位佇列中的每個元素

此範例會呼叫 prime_factors_of 函式來計算其主要因素，以處理 Carmichael 數位佇列中的每個元素。它會使用工作組平行執行這項工作。如需工作組的詳細資訊，請參閱工作平行處理原則。

如果該數字超過四個質素因數，本範例會列印每個 Carmichael 數位的主要因素。

// 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();

範例：已完成的平行容器程序代碼範例

下列程式代碼顯示完整的範例，其會使用平行容器來計算 Carmichael 數位的主要因素。

// 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();
}

此範例會產生下列範例輸出。

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

複製範例程式代碼，並將其貼到 Visual Studio 專案中，或貼到名為 carmichael-primes.cpp 的檔案中，然後在 Visual Studio 命令提示字元視窗中執行下列命令。

cl.exe /EHsc carmichael-primes.cpp

另請參閱

平行容器和物件
 工作平行處理原則
 concurrent_vector 類別
 concurrent_queue 類別
 parallel_invoke函式
 parallel_for函式
 task_group 類別

共用方式為