使用do...while循环，完成一个计算1到100之间6的倍数出现次数的小程序

好的，这是一个关于计算1到100之间6的倍数出现次数的小程序，使用do...while循环实现：

int i = 1, count = 0;
do {
    if (i % 6 == 0) {
        count++;
    }
    i++;
} while (i <= 100);
cout << "1到100之间6的倍数出现了" << count << "次。" << endl;

您可以将此代码复制并粘贴到编译器中，运行程序以获取答案。

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用Java使用do...while循环，完成一个计算1到100之间6的倍数出现次数的小程序

下面是一个用Java实现的计算1到100之间6的倍数出现次数的小程序：

public class Main {
    public static void main(String[] args) {
        int count = 0;
        int i = 1;
        do {
            if (i % 6 == 0) {
                count++;
            }
            i++;
        } while (i <= 100);
        System.out.println("1到100之间6的倍数出现的次数为：" + count);
    }
}

输出结果为：

1到100之间6的倍数出现的次数为：16

在这个程序中，我们使用了do...while循环来遍历1到100之间的整数，如果当前的数字是6的倍数，我们就将计数器count加1。最后，我们输出计数器的值，即为1到100之间6的倍数出现的次数。

用do while循环实现如下功能：(f5-2.c) 设unsigned int uv;从键盘接收2个整数，输出最小公倍数、最大公约数。 请截图展示源程序、公倍数、公约数结果

Sorry, as an AI language model, I am not able to provide screenshots. However, here is the code that solves the problem:

#include <stdio.h>

int main() {
    unsigned int uv, u, v, r, lcm;
    printf("Please enter two positive integers: ");
    scanf("%u %u", &u, &v);
    uv = u * v;
    do {
        r = u % v;
        u = v;
        v = r;
    } while (r);
    lcm = uv / u;
    printf("The greatest common divisor is %u.\n", u);
    printf("The least common multiple is %u.\n", lcm);
    return 0;
}

Explanation:

- We declare and initialize five unsigned integer variables `uv`, `u`, `v`, `r`, and `lcm`.
- We prompt the user to enter two positive integers using the `printf` and `scanf` functions.
- We calculate the product of the two integers `u` and `v` and store it in `uv`.
- We use a `do-while` loop to compute the greatest common divisor (GCD) of `u` and `v` using the Euclidean algorithm. The loop continues as long as the remainder `r` is non-zero. Inside the loop, we update `r` to be the remainder of `u` divided by `v`, `u` to be `v`, and `v` to be `r`.
- Once the `do-while` loop terminates, we compute the least common multiple (LCM) of `u` and `v` by dividing `uv` by `u`.
- Finally, we use the `printf` function to display the GCD and LCM to the user.