在 C 中将整数四舍五入到最接近的十或百 [英] Rounding integers to nearest ten or hundred in C

问题描述

我想用 C 语言编写一个满足以下条件的函数:

I'm trying to think of a function in C that would satisfy the following conditions:

• 它接受一个大于 0 的整数作为参数；
• 它将整数向上舍入到最接近的值，以便只有第一个数字不是零

例如:

53 变成 60..

197 变成 200..

197 comes out as 200..

4937 的结果是 5000..

4937 comes out as 5000..

有没有办法做到这一点，无论尾随零的数量如何，都满足要求?

Is there a way to do this so that the requirement is satisfied regardless of the number of trailing zeroes?

例如，我了解我可以在任何个别情况下如何做到这一点.将 53 除以 10 然后将 ceil() 乘以 10， 但我想要一个可以处理任何值的.

For example, I understand how I could do it in any individual case. divide 53 by 10 then ceil(), multiply by 10, but I would like one that can handle any value.

意见?想法?

推荐答案

没有必要将数字转换为字符串并返回.您可以使用基本的模运算和乘除法来完成此操作.

It's unnecessary to convert the number to a string and back. You can do this using basic modulo arithmetic and multiplication and division.

这是一个纯数字解决方案，希望在运行时间方面更有效:

Here's a pure numeric solution, hopefully somewhat more efficient in terms of running time:

int round_up_to_max_pow10(int n)
{
    int tmp = n;
    int i = 0;
    while ((tmp /= 10) >= 10) {
        i++;
    }

    if (n % (int)(pow(10, i + 1) + 0.5)) {
        tmp++;
    }

    for (; i >= 0; i--) {
        tmp *= 10;
    }

    return tmp;
}

printf("Original: %d; rounded: %d\n", 4937, round_up_to_max_pow10(4937));

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