xglow xglow - 3 months ago 8
C++ Question

Faster Harmonic Number?

I'm a beginner in programming. I am trying to make a program that given two numbers it substracts one harmonic from the other. (Input: n, m / Output: Hn-Hm)

#include <iostream>
#include <iomanip>
#include <math.h>
using namespace std;

int main() {
double n1, n2, h1 = 0, h2 = 0, i; // n = number, h = harmonic
cin >> n1 >> n2;

if (n1 == 0) {
h1 = 0;
}
else {
for (i = 1; i <= n1; i++) {
h1 += 1 / i;
if (i <= n2) {
h2 += 1 / i;
}
}
}
cout << fixed << setprecision(10) << h1 - h2 << endl;

system("pause");
return 0;
}


The program gives correct results but I'm using a website of my university and it says that the program is slow. I've tried to make it faster but I can't figure out how.
Thanks.

Answer

You don't need to calculate the full harmonic numbers. Assuming n1 < n2, the two series will be:

H(n1) = 1 + 1/2 + 1/3 + ... + 1/n1
H(n2) = 1 + 1/2 + 1/3 + ... + 1/n1 + 1/(n1+1) + 1(n1+2) + ... + 1/n2

So when you subtract H(n2) - H(n1), the first n1 terms in the two series cancel each other out, so

H(n2) - H(n1) = 1/(n1+1) + 1(n1+2) + ... + 1/n2

If n1 > n2 the result is the negative of this.

double result = 0, mult = 1;
if (n1 > n2) {
    double temp = n1;
    n1 = n2;
    n2 = temp;
    mult = -1;
}
for (double denom = n1+1; denom <= n2; denom++) {
    result += 1/denom;
}
result *= mult; // Flip the sign if we swapped n1 and n2
cout << fixed << setprecision(10) << result << endl;