junichiro

use async_openai::{Client, types::{CreateCompletionRequestArgs}};

pub struct OpenAi {
    client: Client,
    model: String,
    max_tokens: u16,
}

impl OpenAi {
    pub fn new(model: String, max_tokens: u16) -> Self {

junichiro

use app::openai::OpenAi;

#[tokio::test]
async fn test_openai() {
    let input_text = "Hello, world!";
    let client = OpenAi::new("text-davinci-003".to_string(), 400_u16);
    let response_text = client.ask(input_text).await;
    assert!(
        response_text.contains(input_text),
        "Expected '{}' to contain '{}'",

junichiro

x := x - \frac{\partial}{\partial x}J(x, y) \\
y := y - \frac{\partial}{\partial y}J(x, y) \\

junichiro

err(h_\theta(x), y) = \left\{
\begin{array}{ll}
1 & (h_\theta(x) \geq 0.5, y = 0 \quad or \quad h_\theta(x) \lt 0.5, y = 1) \\
0 & (otherwise)
\end{array}
\right.

junichiro

hx = X * theta;
cost = hx - y;
theta = [0; theta(2:end, :)];
J = ((cost' * cost) + lambda * (theta' * theta)) / (2 * m);

junichiro

a = [1, 2, 3]
b = [2, 3, 4]
ans = 0
for i, v in enumerate(a):
    ans = ans + a[i] * b[i]
print(ans)

junichiro

以下の三角形の頂点から下まで移動するとき, その数値の和の最大値は23になる.

   3
  7 4
 2 4 6
8 5 9 3
この例では 3 + 7 + 4 + 9 = 23.

以下の三角形を頂点から下まで移動するとき, その最大の和を求めよ.

junichiro

import sys


class Problem17:

    COUNT = {
        0: '',
        1: 'one',
        2: 'two',
        3: 'three',

junichiro

import sys


class Problem16:

    def main(self, n):
        return sum(int(i) for i in str(2 ** n))


if __name__ == '__main__':

junichiro

import sys


class Problem15:

    def main(self, n):
        tensor = [list(range(n + 1)) for i in range(n + 1)]
        for i in range(n + 1):
            for j in range(n + 1):
                if i > 0 and j > 0: