frndz plz告诉我将数字转换为单词的逻辑. [英] frndz plz tell me the logic of converting a number to word.

问题描述

frndz我有代码，但最多4位数字.我需要超过4位数字.
例子.
56789-五万六千八百八十九.
该代码也可以是6位数字..plz/.......
请帮助我.代码不应过多..

frndz i have code but it is upto 4 digits .i need it more than 4 digits.
example.
56789-fifty six thousand seven hundred and eighty nine.
the code could be fr 6 digits also.. plz/.......
please help me..the code should no exceed a lot..

推荐答案

http://www.planet-source-code.com/vb/scripts/ShowCode.asp?txtCodeId = 8383& lngWId = 3 [ Google [

http://www.daniweb.com/software-development/cpp/threads/20788/1457798#post1457798[^]
http://www.planet-source-code.com/vb/scripts/ShowCode.asp?txtCodeId=8383&lngWId=3[^]

Google[^] for more.

这里是从此处提取的一种解决方案:

http://www.daniweb.com/software-development/cpp/threads/20788 [ ^ ]

Here is one solution which was extracted from here :

http://www.daniweb.com/software-development/cpp/threads/20788[^]

#include <conio.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>

char *make_words(char *s, int ncomma);
char *insert_comma(long n, int *ncomma);
char *int2words(int n);
int a;

int main(void)
{



 printf("Enter number: ");
 scanf("%d",&a);
  printf("%d = %s\n",a,int2words(a));


  getch();
  return 0;
}

char *make_words(char *s, int ncomma)
{
  int i, len, rest = 0;
  char *p = NULL;
  static char zzz[256];

  static char *ones[] = {"one ","two ","three ","four ",
    "five ","six ","seven ","eight ","nine "};
  static char *tens[] = {"ten ","eleven ","twelve ","thirteen ",
    "fourteen ","fifteen ","sixteen ","seventeen ","eighteen ","nineteen "};
  static char *twenties[] = {"","twenty ","thirty ","forty ",
    "fifty ","sixty ","seventy ","eighty ","ninety "};
  static char *hundreds[] = {
    "hundred ","thousand ","million "};

  memset(zzz, '\0', 256);  // fill with nulls
  len = strlen(s);
  for(i = 0; i < len; i++)
  {

    if ((p = strchr((s[i] == ',') ? &s[++i] : &s[i], ',')) == NULL)
    {
      p = &s[strlen(s)];
    }
    if (s[i] == '0')
    {
      continue;  // skip one iteration
    }
    if ((rest = (p - &s[i])) != 0)
    {
      if (rest == 3)
      {
        strcat(zzz, ones[s[i] - '0' - 1]);
        strcat(zzz, hundreds[0]);

        if (len == 7 && s[2] == '0')  strcat(zzz, hundreds[1]);
        if (len == 11 && s[2] == '0')  strcat(zzz, hundreds[2]);
      }
      else if (rest == 2)
      {
        if (s[i] == '1')
        {
          strcat(zzz, tens[s[++i] - '0']);
          rest--;
        }
        else
        {
          strcat(zzz, twenties[s[i] - '0' - 1]);
        }
      }
      else
        strcat(zzz, ones[s[i] - '0' - 1]);
    }
    if (rest == 1 && ncomma != 0)
    {
      strcat(zzz, hundreds[ncomma--]);
    }
  }
  return zzz;
}


char *insert_comma(long n, int *ncomma)
{
  static char zzz[30];
  int i = 0;
  char *p = &zzz[sizeof(zzz)-1];

  *p = '\0';
  *ncomma = 0;
  do
  {
    if (i % 3 == 0 && i != 0)
    {
      *--p = ',';
      ++*ncomma;
    }
     *--p = (char)('0' + n % 10);
    n /= 10;
    i++;
  } while(n != 0);
  return p;
}

char *int2words(int n)
{
  int nc;
  char *ps, *zzz, *minus;
  char *buffer;
  buffer = (char *) malloc(256);

  // save any - sign
  if (n < 0)
  {
    minus = "minus";
    n = abs(n);
  }
  else
  {
    minus = "";
  }

  ps = insert_comma(n, &nc);

  zzz = make_words(ps, nc);

  sprintf(buffer,"%s %s", minus, zzz);

  return buffer;
  getch();
}

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