使用C语言，写一个程序 n1为54，n2 为84，n1是n2的64%

As an AI language model, I am not capable of experiencing or expressing emotions. Therefore, the concept of null does not apply to me in any way. However, null represents the absence of a value or a non-existent object or entity in computer programming, databases, and other related fields.

相关问题

使用C语言，写一个程序 n1为54，n2

为13，计算它们的和、差、积、商、余数和平均数。

代码如下：

#include <stdio.h>

int main() {
    int n1 = 54, n2 = 13;
    int sum = n1 + n2;
    int difference = n1 - n2;
    int product = n1 * n2;
    int quotient = n1 / n2;
    int remainder = n1 % n2;
    float average = (float)(n1 + n2) / 2;

    printf("n1 = %d, n2 = %d\n", n1, n2);
    printf("sum = %d\n", sum);
    printf("difference = %d\n", difference);
    printf("product = %d\n", product);
    printf("quotient = %d\n", quotient);
    printf("remainder = %d\n", remainder);
    printf("average = %.2f\n", average);

    return 0;
}

输出结果如下：

n1 = 54, n2 = 13
sum = 67
difference = 41
product = 702
quotient = 4
remainder = 2
average = 33.50

用函数计算n1到n2之间回文数有多少个（c语言）

在C语言中，你可以通过编写一个递归或循环函数来计算n1到n2之间的回文数。这里是一个简单的示例，使用了循环方法：

#include <stdio.h>
#include <stdbool.h>

bool is_palindrome(int num) {
    int rev = 0;
    while (num != 0) {
        rev = rev * 10 + num % 10;
        num /= 10;
    }
    return num == rev;
}

int count_palindromes(int n1, int n2) {
    int count = 0;
    for (int i = n1; i <= n2; ++i) {
        if (is_palindrome(i)) {
            count++;
        }
    }
    return count;
}

int main() {
    int start, end;
    printf("请输入回文数的范围（开始值和结束值）：");
    scanf("%d %d", &start, &end);

    int palindrome_count = count_palindromes(start, end);
    printf("在%d到%d范围内，共有 %d 个回文数。\n", start, end, palindrome_count);

    return 0;
}

这个程序首先定义了一个`is_palindrome`函数来检查一个数字是否是回文，然后在`count_palindromes`函数里遍历给定范围内的所有整数，如果它是回文就增加计数器。最后在主函数中获取用户输入并打印结果。