Write a recursive function to calculate sum of first n natural numbers.

要编写一个递归函数来计算前n个自然数的和，可以参考以下Python代码：

def sum_of_natural_numbers(n):
    """
    使用递归计算前n个自然数的和
    参数:
    n -- 需要计算和的自然数的数量
    返回:
    和 -- 前n个自然数的总和
    """
    if n == 0:  # 基线条件：当n为0时，返回0（空和）
        return 0
    else:
        # 递归条件：如果n不为0，则将当前数字n加上前一个数的和
        return n + sum_of_natural_numbers(n - 1)  # 召唤自身并传入n-1

# 测试函数
n = 5
result = sum_of_natural_numbers(n)
print(f"The sum of the first {n} natural numbers is: {result}")

这个递归函数首先检查基本情况（即n等于0），在这种情况下直接返回0。对于其他的n值，函数会通过递归调用自身，每次减少1，直到达到基础情况为止。在每次递归调用过程中，都会把当前的数字n加到前一个数的和上。

相关问题

Write a program in C to calculate the sum of numbers from 1 to n using recursion. Input: Input a positve integer n. Output : Compute the summation of 1+2+...+n. Sample of Input: 5 Sample of Output: 15

Sure, I can help you with that. Here's the C program to calculate the sum of numbers from 1 to n using recursion:

#include <stdio.h>

int sum(int n);

int main() {
    int n, result;
    printf("Enter a positive integer: ");
    scanf("%d", &n);
    result = sum(n);
    printf("The sum of numbers from 1 to %d is %d.\n", n, result);
    return 0;
}

int sum(int n) {
    if (n == 0) {
        return 0;
    } else {
        return n + sum(n-1);
    }
}

Explanation:

The program takes a positive integer `n` as input from the user and calls the `sum()` function with `n` as argument.

The `sum()` function is a recursive function that takes an integer `n` as input and returns the sum of numbers from 1 to `n`.

If `n` is 0, the function returns 0. Otherwise, it adds `n` to the sum of numbers from 1 to `n-1` by calling itself recursively with `n-1` as argument.

Finally, the program outputs the result, which is the sum of numbers from 1 to `n`.