Question

Modify the sum_thread.cpp program to compute harmonic sum = 1 + 1/2 + 1/3 + 1/4 + … 1/n.

Let’s estimate natural using n = 20.

sum_thread.cpp

#include <chrono>
#include <iostream>
#include <mutex>
#include <random>
#include <utility>
#include <vector>
#include <thread>

using namespace std;

constexpr long long size= 1000000;   
mutex myMutex;

void sumUp(unsigned long long& sum, const vector<int>& val, 
   unsigned long long beg, unsigned long long end){
   long long localSum = 0;
    for (auto it= beg; it < end; ++it){
        localSum+= val[it];
    }
    lock_guard<mutex> myLock(myMutex);
    sum += localSum;
}

int main(){

  cout << endl;

  vector<int> randValues;
  randValues.reserve(size);

  mt19937 engine (0);
  uniform_int_distribution<> uniformDist(1,10);
  for ( long long i=0 ; i< size ; ++i)
     randValues.push_back(uniformDist(engine));
 
  unsigned long long sum= 0;
  auto start = chrono::system_clock::now();

  int threads = 4;
  thread t[threads];
  long long slice = size / threads;
  int startIdx=0;
  for (int i = 0; i < threads; ++i) {
    cout << "Thread[" << i << "] - slice ["
         << startIdx << ":" << startIdx+slice-1 << "]" << endl;
    t[i] = thread(sumUp, ref(sum), ref(randValues), startIdx, startIdx+slice-1);
    startIdx += slice;
  }

  for (int i = 0; i < threads; ++i)
     t[i].join();

  chrono::duration<double> dur= chrono::system_clock::now() - start;
  cout << "Time for addition " << dur.count() << " seconds" << endl;
  cout << "Result: " << sum << endl;

  cout << endl;

}

Answer 1

// C++ program to find sum of harmonic series

#include<bits/stdc++.h>

using namespace std;   

// Function to return sum of harmonic series

double sum(int n)

{

  double i, s = 0.0;

  for(i = 1; i <= n; i++)

      s = s + 1 / i;

  return s;

}

// Driver code

int main()

{

    int n = 5;      

    cout << "Sum is " << sum(n);

    return 0;

}

You can also use this code:

#include<iostream>
using namespace std;
int main()
{
long d,n,a;
long an;
cout<<"Enter first term : ";
cin>>a;
cout<<"Enter difference : ";
cin>>d;
cout<<"Enter number of terms :";
cin>>n;
cout<<"Harmonic progression : ";
for(int y=1;y<=n;y++)
{
an=a+(y-1)*d;
cout<<"1/"<<an<<" ";
}
return 0;
}

Also Use this code:

Harmonic progression is a series whose inverse will be an arithmetic progression. i.e. if for a harmonic progression A1, A2, A3.. An, there is an arithmetic
progression 1/A1, 1/A2, 1/A3.

So, a general HP is :
1/a, 1/(a+d), 1/(a+2d), … 1/(a + nd),where 1/a is the first term and d is the common difference of the reversed AP.

Input:a = 3, d = 2, n = 5
Output:The sum of HP is 0.878211
Execution:Sum = ⅓ + ⅕ + 1/7 + 1/9 + 1/11 = 0.878211

Code:
#include <iostream>
using namespace std;
float findSeriesSum(int a, int d, int n){
float sumVal = 0;
float term = 0;
for(float i = 1; i <= n; i++){
term = (1.0)/(float)(a + (i-1)*d);
sumVal += term;
}
return sumVal;
}
int main(){
int n, a, d ;
cin>>n>>a>>d;
cout<<"The sum of HP is "<<findSeriesSum(a, d, n);
return 0;
}