我正在尝试使用opemMP做一个“谐波进展总和”问题的并行版本。 但输出依赖于输入而不同。 (并行和串行)谐波进展总和C++ openMP
程序:
#include "stdafx.h"
#include <iostream>
#include <sstream>
#include <omp.h>
#include <time.h>
#define d 10 //Numbers of Digits (Example: 5 => 0,xxxxx)
#define n 1000 //Value of N (Example: 5 => 1/1 + 1/2 + 1/3 + 1/4 + 1/5)
using namespace std;
void HPSSeguencial(char* output) {
long unsigned int digits[d + 11];
for (int digit = 0; digit < d + 11; ++digit)
digits[digit] = 0;
for (int i = 1; i <= n; ++i) {
long unsigned int remainder = 1;
for (long unsigned int digit = 0; digit < d + 11 && remainder; ++digit) {
long unsigned int div = remainder/i;
long unsigned int mod = remainder % i;
digits[digit] += div;
remainder = mod * 10;
}
}
for (int i = d + 11 - 1; i > 0; --i) {
digits[i - 1] += digits[i]/10;
digits[i] %= 10;
}
if (digits[d + 1] >= 5) {
++digits[d];
}
for (int i = d; i > 0; --i) {
digits[i - 1] += digits[i]/10;
digits[i] %= 10;
}
stringstream stringstreamA;
stringstreamA << digits[0] << ",";
for (int i = 1; i <= d; ++i) {
stringstreamA << digits[i];
}
string stringA = stringstreamA.str();
stringA.copy(output, stringA.size());
}
void HPSParallel(char* output) {
long unsigned int digits[d + 11];
for (int digit = 0; digit < d + 11; ++digit)
digits[digit] = 0;
int i;
long unsigned int digit;
long unsigned int remainder;
#pragma omp parallel for private(i, remainder, digit)
for (i = 1; i <= n; ++i) {
remainder = 1;
for (digit = 0; digit < d + 11 && remainder; ++digit) {
long unsigned int div = remainder/i;
long unsigned int mod = remainder % i;
digits[digit] += div;
remainder = mod * 10;
}
}
for (int i = d + 11 - 1; i > 0; --i) {
digits[i - 1] += digits[i]/10;
digits[i] %= 10;
}
if (digits[d + 1] >= 5) {
++digits[d];
}
for (int i = d; i > 0; --i) {
digits[i - 1] += digits[i]/10;
digits[i] %= 10;
}
stringstream stringstreamA;
stringstreamA << digits[0] << ",";
for (int i = 1; i <= d; ++i) {
stringstreamA << digits[i];
}
string stringA = stringstreamA.str();
stringA.copy(output, stringA.size());
}
int main() {
//Sequential Method
cout << "Sequential Method: " << endl;
char outputSeguencial[d + 10];
HPSSeguencial(outputSeguencial);
cout << outputSeguencial << endl;
//Cleaning vector
string stringA = "";
stringA.copy(outputSeguencial, stringA.size());
//Parallel Method
cout << "Parallel Method: " << endl;
char outputParallel[d + 10];
HPSParallel(outputParallel);
cout << outputParallel << endl;
system("PAUSE");
return 0;
}
实例:
输入:
#define d 10
#define n 1000
输出:
Sequential Method:
7,4854708606╠╠╠╠╠╠╠╠╠╠╠╠
Parallel Method:
6,6631705861╠╠╠╠╠╠╠╠╠╠╠╠╠╠╠╠ÇJ^
输入:
#define d 12
#define n 7
输出:
Sequential Method:
2,592857142857╠╠╠╠╠╠╠╠╠╠╠╠╠╠ÀÂ♂ü─¨@
Parallel Method:
2,592857142857╠╠╠╠╠╠╠╠╠╠╠╠╠╠╠╠╠╠ÇJJ
个问候
Pastecode更新digits
阵列时
http://pastecode.org/index.php/view/62768285
一些编译器(阅读GCC)在x86上用'lock addl'实现原子'+ ='。关键部分通过互斥体实现,包括运行时函数调用,因此比原子方法慢。实际上,由于内部循环是规则的,所以线程会在几次迭代之后延迟同步,并且不会有等待。缓存一致性会损害并行增益。 –