出一道c语言条件判断的难题，只用if和switch解答，不用其他语句

这是一个用C语言仅限if和switch完成的挑战题目，目标是判断一个整数属于特定的范围，并执行相应的操作：

#include <stdio.h>

int main() {
    int num;
    
    printf("Enter an integer: ");
    scanf("%d", &num);

    // 使用if-else if-else链路
    if (num >= 0 && num <= 5) {
        printf("Number is between 0 and 5. Operation A.\n");
    } else if (num > 5 && num <= 10) {
        printf("Number is between 6 and 10. Operation B.\n");
    } else if (num > 10 && num <= 20) {
        printf("Number is between 11 and 20. Operation C.\n");
    } else if (num > 20) {
        printf("Number is greater than 20. Operation D.\n");
    } else { // 如果不在上述范围内
        printf("Number is less than 0 or not within the given range.\n");
    }

    // 使用switch-case
    switch (num / 5) {
        case 0:
            printf("Number is a multiple of 5 in range [0-5].\n");
            break;
        case 1:
            printf("Number is either 6, 11 or higher multiples of 5.\n");
            break;
        default:
            printf("Number does not fall into the above categories.\n");
            break;
    }

    return 0;
}

在这个例子中，我们首先用if-else if-else检查数字是否在每个区间内，然后用switch-case将数字分成每5的倍数段进行判断。