Can you ever hope to be an actually great programmer if you can't solve Math Olympiad questions?

UFOs Are A Psyop Shirt $21.68 |

Skip to content
# Can you ever hope to be an actually great programmer if you can't solve Math Olympiad questions?

###

Can you ever hope to be an actually great programmer if you can't solve Math Olympiad questions?

UFOs Are A Psyop Shirt $21.68 |

DevinAI was a scam by chinkoid nerds.

It's a serious question. How do you objectively assess your problem solving skills. If you can't do this, what can you do?

Import libraries other people have written to solve the problems for you.

with Informatics Olympiad questions, you moron

When will the lynching of Devin chinks begin? I want to see them being naked and humiliated at the streets of San Francisco.

Do you have space left for Olympiad ideas too to go with it too? No? Then frick off.

Can you ever hope to actually reproduce if you can solve Math Olympiad questions?

math has no relation to computing

moron

pseud.

None of the programmers that are regarded as genius programmers are math olympiad gays, afaik. John Carmack barely knows 2+2=4.

True.

t. javascript developer

>math has no relation to computing

that's about of a stretch. that said most programmers won't need anything more than basic algebra.

>wanting to be a great programmer

Here's my shitty ass solution

#include <iostream>

#include <cmath>

using namespace std;

int lengthOfTriangles[] = { }; // Insert some number here that satisfies the inequality

int main() {

int n = sizeof(lengthOfTriangles) / sizeof(lengthOfTriangles[0]);

int totalSum = 0;

double totalSumReciprocal = 0.0;

for (int i = 0; i < n; i++) {

totalSum += lengthOfTriangles[i];

totalSumReciprocal += 1.0 / lengthOfTriangles[i];

}

// t_1,t_2,...,t_n must pass the following inequality

if (pow(n, 2) + 1 <= totalSum * totalSumReciprocal) {

cout << "The given inequality does not hold." << endl;

return 0;

}

// moronic ass O(n^3) solution

for (int i = 0; i < n - 2; i++) {

for (int j = i + 1; j < n - 1; j++) {

for (int k = j + 1; k < n; k++) {

bool cond1 = lengthOfTriangles[i] + lengthOfTriangles[j] > lengthOfTriangles[k];

bool cond2 = lengthOfTriangles[j] + lengthOfTriangles[k] > lengthOfTriangles[i];

bool cond3 = lengthOfTriangles[k] + lengthOfTriangles[i] > lengthOfTriangles[j];

if (!cond1 || !cond2 || !cond3) {

cout << "homie what?" << endl;

return 0;

}

}

}

}

cout << "All conditions are satisfied. QED." << endl;

return 0;

}

Welcome to shit on my code

>prove

>here's my code

that's not how that works, ma homie.

True. A formal proof that generalizes to all cases is the right proof in the context of a Math Olympiad. Now while we're at it, let me continue to listen to sissy twink hypno.

i feel like there's a propriety of triangles you have to prove here and i can't remember it but if i could i would solve this in no time

If you want to cheat, here's the solution

Well, to prove this, you can use the triangle inequality to prove that they form a triangle if and only if the following 3 relations hold

$t_i leq t_j + t_k$

$t_j leq t_i + t_k$

$t_k leq t_i + t_j$

For any tuple $(t_i, t_j, t_k)$

To get the implications you'd have to do some algebraic manipulations and maybe use the set of inequalities between the harmonic, arithmetic, quadratic, and geometric means

Possibly some other inequalities as well like Cauchy-Scwarz.

t.mathgay who is filling his last year for the second time