【Code Llama】最強コード生成AIの使い方と実践を解説

!pip install git+https://github.com/huggingface/transformers.git@main accelerate

from transformers import AutoTokenizer
import transformers
import torch

# モデルとトークナイザーの準備
model = "codellama/CodeLlama-7b-hf"
tokenizer = AutoTokenizer.from_pretrained(model)
pipeline = transformers.pipeline(
    "text-generation",
    model=model,
    torch_dtype=torch.float16,
    device_map="auto",
)

sequences = pipeline(
    'ここにプロンプトを入力します',
    do_sample=True,
    top_k=10,
    temperature=0.1,
    top_p=0.95,
    num_return_sequences=1,
    eos_token_id=tokenizer.eos_token_id,
    max_length=200,
)
for seq in sequences:
    print(f"Result: {seq['generated_text']}")

## 解答
```python
def bubble_sort(arr):
    for i in range(len(arr)):
        for j in range(len(arr) - i - 1):
            if arr[j] > arr[j + 1]:
                arr[j], arr[j + 1] = arr[j + 1], arr[j]
    return arr


if __name__ == '__main__':
    arr = [1, 5, 2, 4, 3]
    print(bubble_sort(arr))
```

この問題においては、与えられた整数Nに対し、
フィボナッチ数列のN番目の数値を返すプログラムを作成します。 
ここでフィボナッチ数列とは、前の二つの数の和が次の数になる数列のことを指します。

# In this problem, you are required to create a program 
# that returns the N-th number of the Fibonacci sequence for a given integer N. 
# The Fibonacci sequence is a series of numbers 
# in which each number is the sum of the two preceding ones.
# Input Format: integer
# Output Format: integer

def

def fib(n):
    if n == 0:
        return 0
    elif n == 1:
        return 1
    else:
        return fib(n - 1) + fib(n - 2)


if __name__ == "__main__":
    n = int(input())
    print(fib(n))

def fibonacci(n):
    if n == 0:
        return 0
    elif n == 1:
        return 1
    else:
        return fibonacci(n - 1) + fibonacci(n - 2)


n = int(input())
print(fibonacci(n))

この問題では、1からNまでの数を並び替えて、
その要素の積を「prod」と、要素の和を「sum」とします。
与えられた正の整数Nに対して、「prod」が「sum」で割り切れるかどうかを判断する問題です。
入力はテストケース数と各テストケースのNの値で、出力は「YEAH」か「NAH」です。

# This problem is asked in the HackerEarth contest.
# Problem Statement: Consider a permutation of numbers from 1 to N written on a paper. Let’s denote the product of its element as ‘prod’ and the sum of its elements as ‘sum’. Given a positive integer N, your task is to determine whether ‘prod’ is divisible by ‘sum’ or not.
# Input Format: First input will be an integer T.  It depicts a number of test cases. Followed by the value for each test case. Each test case will contain an integer N (1<= N <=10^9). It is nothing but the length of the permutation.
# Output Format: For each test case, print “YEAH” if ‘prod’ is divisible by ‘sum’, otherwise print “NAH”

def

def check_divisibility(n):
    prod = 1
    sum = 0
    for i in range(1, n+1):
        prod *= i
        sum += i
    if prod % sum == 0:
        return "YEAH"
    else:
        return "NAH"


t = int(input())
for _ in range(t):
    n = int(input())
    print(check_divisibility(n))

def isDivisible(n):
    sum = 0
    while n > 0:
        sum = sum + (n % 10)
        n = n // 10
    if sum % 3 == 0:
        return True
    else:
        return False


n = 123
if isDivisible(n):
    print("YEAH")
else:
    print("NAH")

与えられた文字列(input関数を用いて)から重複する文字をすべて削除します。
最終的な出力文字列には、各文字が1回だけ含まれます。
文字列内の各文字の順序はそのままである必要があります。
文字列を一度だけ走査できます。

# You have given a string (from input function). You need to remove all the duplicates from the string.
# The final output string should contain each character only once. The respective order of the characters inside the string should remain the same.
# You can only traverse the string at once.
# Input format: input
# Output format: one string

def 

def remove_duplicates(string):
    new_string = ""
    for i in range(len(string)):
        if string[i] not in new_string:
            new_string += string[i]
    return new_string

if __name__ == '__main__':
    string = input()
    print(remove_duplicates(string))

def remove_duplicates(string):
    string = list(string)
    for i in range(len(string)):
        if string[i] in string[i + 1 :]:
            string.remove(string[i])
    return "".join(string)


# Test
print(remove_duplicates("geeksforgeeks"))