Write a function that takes an integer n that is greater than 1 and returns the largest integer that is smaller than n and evenly divides n.

编写一个函数，接收一个大于1的整数n，该函数的目标是找到能被n整除的最大整数因子。这个因子应该小于n本身。这通常用于求解最大公约数（Greatest Common Divisor, GCD），尤其是当需要找到两个数的最大公约数时，可以取较小的那个数作为输入。

在Python中，可以使用math库中的gcd函数来直接计算最大公约数，但如果我们要手动实现这个功能，可以使用欧几里得算法（Euclidean Algorithm）：

def largest_divisor(n):
    if n <= 1:
        raise ValueError("Input must be greater than 1")
    divisor = 1
    for i in range(2, int(n**0.5) + 1):  # 遍历到根号n即可，因为较大因子必定有个对应的小于根号n
        if n % i == 0:
            divisor = i  # 更新当前找到的最大因子
            break
    return divisor

如果你需要创建一个自定义版本来理解其工作原理，上述代码就是基础。它通过循环遍历n的所有可能因数，并在找到第一个因子时停止，返回那个因子。

相关问题

use C program Write a function digit(n, k) that returns the k th digit (from the right) in n (a positive integer). For example digit(829, 1) returns 9, digit(829, 2) returns 2 and digit(829, 3) returns 8. if k is greater than the number of digits in n , have the function return 0.

Sure, here's a C program that defines a function `digit(n, k)` that returns the `k`th digit (from the right) in `n` (a positive integer):

#include <stdio.h>

int digit(int n, int k) {
    int count = 0;
    while (n > 0) {
        count++;
        if (count == k) {
            return n % 10;
        }
        n /= 10;
    }
    return 0;
}

int main() {
    int n = 829;
    int k = 1;
    printf("The %dth digit (from the right) in %d is %d\n", k, n, digit(n, k));
    
    k = 2;
    printf("The %dth digit (from the right) in %d is %d\n", k, n, digit(n, k));
    
    k = 3;
    printf("The %dth digit (from the right) in %d is %d\n", k, n, digit(n, k));
    
    k = 4;
    printf("The %dth digit (from the right) in %d is %d\n", k, n, digit(n, k));
    
    return 0;
}

This program defines a function `digit(n, k)` that takes two arguments: `n` (the positive integer) and `k` (the index of the digit to return, from the right). The function works by counting the number of digits in `n` as it repeatedly divides `n` by 10. If `count` equals `k`, the function returns the last digit of `n`. If `count` exceeds the number of digits in `n`, the function returns 0.

I hope that helps! Let me know if you have any further questions.

Write a function, is_prime, that takes a single integer argument and returns True when the argument is a prime number and False otherwise.

Sure, here's the code for the is_prime function in Python:

def is_prime(n):
    if n <= 1:
        return False
    for i in range(2, int(n**0.5)+1):
        if n % i == 0:
            return False
    return True

This function checks if the input integer is greater than 1 (since 1 is not a prime number), and then iterates through all integers from 2 to the square root of the input integer. If any of these integers divide the input integer evenly, then it's not a prime and the function returns False. Otherwise, the input integer is a prime and the function returns True.