如何将整数值转换为罗马数字字符串？

Question

如何将整数转换为C中罗马数字的String表示？

Answer 1

最简单的方法可能是为复杂的情况设置三个数组,并使用一个简单的函数,如:

// convertToRoman:
//   In:  val: value to convert.
//        res: buffer to hold result.
//   Out: n/a
//   Cav: caller responsible for buffer size.

void convertToRoman (unsigned int val, char *res) {
    char *huns[] = {"", "C", "CC", "CCC", "CD", "D", "DC", "DCC", "DCCC", "CM"};
    char *tens[] = {"", "X", "XX", "XXX", "XL", "L", "LX", "LXX", "LXXX", "XC"};
    char *ones[] = {"", "I", "II", "III", "IV", "V", "VI", "VII", "VIII", "IX"};
    int   size[] = { 0,   1,    2,     3,    2,   1,    2,     3,      4,    2};

    //  Add 'M' until we drop below 1000.

    while (val >= 1000) {
        *res++ = 'M';
        val -= 1000;
    }

    // Add each of the correct elements, adjusting as we go.

    strcpy (res, huns[val/100]); res += size[val/100]; val = val % 100;
    strcpy (res, tens[val/10]);  res += size[val/10];  val = val % 10;
    strcpy (res, ones[val]);     res += size[val];

    // Finish string off.

    *res = '\0';
}

这将处理任何无符号整数,尽管大数字M在前面会有很多字符,并且调用者必须确保它们的缓冲区足够大.

一旦数量减少到1000以下,它就是一个简单的3表查找,每个查找数百,数十和单位.例如,取其中的情况val是314.

val/100会3在这种情况下,这样的huns阵列查找会给CCC,然后val = val % 100给你14的tens查找.

然后val/10会1在这种情况下,这样的tens阵列查找会给X,然后val = val % 10给你4的ones查找.

那么val将4在那种情况下,所以ones数组查找将给出IV.

这给你CCCXIV的314.

缓冲区溢出检查版本是一个简单的步骤:

// convertToRoman:
//   In:  val: value to convert.
//        res: buffer to hold result.
//   Out: returns 0 if not enough space, else 1.
//   Cav: n/a

int convertToRoman (unsigned int val, char *res, size_t sz) {
    char *huns[] = {"", "C", "CC", "CCC", "CD", "D", "DC", "DCC", "DCCC", "CM"};
    char *tens[] = {"", "X", "XX", "XXX", "XL", "L", "LX", "LXX", "LXXX", "XC"};
    char *ones[] = {"", "I", "II", "III", "IV", "V", "VI", "VII", "VIII", "IX"};
    int   size[] = { 0,   1,    2,     3,    2,   1,    2,     3,      4,    2};

    //  Add 'M' until we drop below 1000.

    while (val >= 1000) {
        if (sz-- < 1) return 0;
        *res++ = 'M';
        val -= 1000;
    }

    // Add each of the correct elements, adjusting as we go.

    if (sz < size[val/100]) return 0;
    sz -= size[val/100];
    strcpy (res, huns[val/100]);
    res += size[val/100];
    val = val % 100;

    if (sz < size[val/10]) return 0;
    sz -= size[val/10];
    strcpy (res, tens[val/10]);
    res += size[val/10];
    val = val % 10;

    if (sz < size[val) return 0;
    sz -= size[val];
    strcpy (res, ones[val]);
    res += size[val];

    // Finish string off.

    if (sz < 1) return 0;
    *res = '\0';
    return 1;
}

虽然,在那时,你可以考虑将数百,数十和单位的处理重构成一个单独的函数,因为它们非常相似.我会把它作为额外的练习.