22

如何将整数转换为 C 中罗马数字的字符串表示形式？

4

3 回答 3

37

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，例如：

// 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，尽管大数字在前面会有很多M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，百、十和单位各一个。例如，以valis为例314。

val/100将3在这种情况下，因此huns数组查找将给出CCC，然后val = val % 100为您14提供tens查找。

然后val/10将1在这种情况下，因此tens数组查找将给出X，然后val = val % 10为您4提供ones查找。

然后val将4在这种情况下，因此ones数组查找将给出IV.

那给你CCCXIV。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;
}

尽管在那时，您可以考虑将数百、十和单位的处理重构为一个单独的函数，因为它们非常相似。我会把它作为一个额外的练习。

于 2011-02-13T20:22:01.313 回答
3

对于困难的情况，不要使用娘娘腔预先计算好的地图。

/* roman.c */
#include <stdio.h>

/* LH(1) roman numeral conversion */
int RN_LH1 (char *buf, const size_t maxlen, int n)
{
  int S[]  = {    0,   2,   4,   2,   4,   2,   4 };
  int D[]  = { 1000, 500, 100,  50,  10,   5,   1 };
  char C[] = {  'M', 'D', 'C', 'L', 'X', 'V', 'I' };
  const size_t L = sizeof(D) / sizeof(int) - 1;
  size_t k = 0; /* index into output buffer */
  int i = 0; /* index into maps */
  int r, r2;

  while (n > 0) {
    if (D[i] <= n) {
      r = n / D[i];
      n = n - (r * D[i]);
      /* lookahead */
      r2 = n / D[i+1];
      if (i < L && r2 >= S[i+1]) {
        /* will violate repeat boundary on next pass */
        n = n - (r2 * D[i+1]);
        if (k < maxlen) buf[k++] = C[i+1];
        if (k < maxlen) buf[k++] = C[i-1];
      }
      else if (S[i] && r >= S[i]) {
        /* violated repeat boundary on this pass */
        if (k < maxlen) buf[k++] = C[i];
        if (k < maxlen) buf[k++] = C[i-1];
      }
      else
        while (r-- > 0 && k < maxlen)
          buf[k++] = C[i];
    }
    i++;
  }
  if (k < maxlen) buf[k] = '\0';
  return k;
}

/* gcc -Wall -ansi roman.c */
int main (int argc, char **argv)
{
  char buf[1024] = {'\0'};
  size_t len;
  int k;
  for (k = 1991; k < 2047; k++)
  {
    len = RN_LH1(buf, 1023, k);
    printf("%3lu % 4d %s\n", len, k, buf);
  }
  return 0;
}

您实际上也不需要声明S。应该很容易理解为什么。

于 2017-05-23T23:24:31.007 回答
-2 
    static string ConvertToRoman(int num)
    {
        int d = 0;
        string result = "";
        while (num > 0)
        {
            int n = num % 10;
            result = DigitToRoman(n, d) + result;
            d++;
            num = num / 10;
        }
        return result;
    }
    static string DigitToRoman(int n, int d)
    {
        string[,] map = new string[3, 3] { { "I", "V", "X" }, { "X", "L", "C" }, { "C", "D", "M" } };
        string result="";
        if (d <= 2)
        {
            switch (n)
            {
                case 0:
                    result = "";
                    break;
                case 1:
                    result = map[d, 0];
                    break;
                case 2:
                    result = map[d, 0] + map[d, 0];
                    break;
                case 3:
                    result = map[d, 0] + map[d, 0] + map[d, 0];
                    break;
                case 4:
                    result = map[d, 0] + map[d, 1];
                    break;
                case 5:
                    result = map[d, 1];
                    break;
                case 6:
                    result = map[d, 1] + map[d, 0];
                    break;
                case 7:
                    result = map[d, 1] + map[d, 0] + map[d, 0];
                    break;
                case 8:
                    result = map[d, 1] + map[d, 0] + map[d, 0] + map[d, 0];
                    break;
                case 9:
                    result = map[d, 0] + map[d, 2];
                    break;
            }
        }
        else if (d == 3 && n < 5)
        {
            while (--n >= 0)
            {
                result += "M";
            }
        }
        else
        {
            return "Error! Can't convert numbers larger than 4999.";
        }
        return result;
    }
于 2015-07-28T19:32:56.463 回答