Calculating Pi

Alright, for the life of me, I can't figure out what is wrong with my code. I'm trying to use this formula to calculate pi in c++. I've tried it with a while loop too, but it always gives me pi = 1. Can anyone help me out?

#include <iomanip>
#include <iostream>

using namespace std;

int main()
{
    char reply;

    double pi=0;
    long n;
    long i;

    cout << "Enter the value of n: ";
    cin >> n;
   
    // Set the state of the output stream to format using 2 places of decimals.
    cout << setprecision(4);
    cout << fixed << setprecision(2);// Set the output stream to display floating point numbers in fixed point notation


for (i = 0; i <= n; i++) {
    if (i%2 == 0)
        pi = (pi+(1 / (2 * i + 1)));
    else
        pi = pi-(1 / (2 * i + 1));
    cout << pi << endl;
}
pi = pi * 4;
cout << pi;

// "Pause" so the user can see the results

    cout << "\nEnter q to quit... ";
    cin >> reply;
return 0;

}

I ran your program. It only displayed:
kill me
kill me
kill me
please kill me

Wow! Is it really you?

so this is the Gregory–Leibniz series?

i bet it would work if you used C....

still, i would try if ((i%2) == 0), don't know if that's it, but it might be..

I remember calculating Pi by directly drawing a circle , using pythagoras theorem, then just counting pixels and dividing their number by radius.

>>1
i&1 is 35% more efficient than i%2

lazy mode

ji = 0;

for (iter = 0:2:10)

  ji = ji + (1 / (2 * iter + 1));

  ji = ji - (1 / (2 * (iter + 1) + 1);

  endfor;

ji = ji * 4;

Jim Profit

If you have posted in this thread YHBT!

>>10 nah, i learnt something =)

similar but faster..

function ji = pivalX(n, n2 = 0)

longrange = [n2:n];

lrplus = 1 ./ (longrange .* 4 .+ 1);

lrminus = 1 ./ (longrange .* 4 .+ 3);

ji = sum(lrplus .- lrminus);
 
ji = ji * 4;

endfunction;

>pivalX(10,000,000)
>elapsed time is 1.4976
>ans = 3.14159260358980

pi += (1.0 / (2.0 * i + 1.0));
    else
pi -= (1.0 / (2.0 * i + 1.0));

maybe

so why does this approximate pi though?

>>11
Oh fuck, >>1 is the 13yo retard-kun!

>>14
the jew formulas say so apparently, what are you a nazi

#define PI 22.0/7.0

>>14
taylor series of arctan(x) around x=1

I can't figure out what is wrong with my code

cepples

also the true men calculate pi with monte-carlo

>>17
Terrible!

#include "stdio.h"

long double GetPI(void){
long double pi;
_asm("fldpi;fstpt %pi")
return pi;
}

int main(void){
printf("%.20Lg\n",GetPI());
return 0;
}