用c语言自行编写代码实现求n从1-100时，n的和，1/n的和，素数的个数。

在C语言中，我们可以分别编写函数来计算给定范围内的整数之和、倒数之和以及素数的数量。这里是一个简单的实现示例：

#include <stdio.h>
#include <math.h>

// 计算数组元素之和
double sum(int arr[], int n) {
    double total = 0;
    for (int i = 0; i < n; i++) {
        total += arr[i];
    }
    return total;
}

// 计算数组元素倒数之和，避免除以零的错误，这里假设n的范围不会包含0
double reciprocal_sum(double arr[], int n) {
    double total = 0.0;
    for (int i = 1; i <= n; i++) {
        if (arr[i - 1] != 0)
            total += 1 / arr[i - 1];
    }
    return total;
}

// 判断是否为素数
int is_prime(int num) {
    if (num <= 1) return 0;
    if (num == 2 || num == 3) return 1;
    if (num % 2 == 0 || num % 3 == 0) return 0;

    for (int i = 5; i * i <= num; i += 6) {
        if (num % i == 0 || num % (i + 2) == 0)
            return 0;
    }
    return 1;
}

int main() {
    int n = 100;
    int numbers[n]; // 存放1到n的整数
    double reciprocals[n]; // 存放1到n的倒数

    // 初始化数组
    for (int i = 0; i < n; i++) {
        numbers[i] = i + 1;
        reciprocals[i] = 1.0 / (i + 1);
    }

    // 计算和与倒数和
    double sum_of_numbers = sum(numbers, n);
    double sum_of_reciprocals = reciprocal_sum(reciprocals, n);

    // 计算素数数量
    int prime_count = 0;
    for (int i = 2; i <= n; i++) {
        if (is_prime(i)) prime_count++;
    }

    printf("Sum of numbers from 1 to 100: %.2f\n", sum_of_numbers);
    printf("Sum of reciprocals from 1 to 100 (excluding 0): %.2f\n", sum_of_reciprocals);
    printf("Number of primes between 1 and 100: %d\n", prime_count);

    return 0;
}