杂记：阿拉伯数字转中文大写数字

　　之前面试过程中碰到一道笔试题，要求将阿拉伯数字的金额转换成中文大写数字，类似发票上的写法。当时只是简单的将小数点之前的数字转换为汉字，并没有实现小数点之后的数字以及当数字中间出现多个0的情况。

　　现将完善后的代码记录下来，不足之处还望各位不吝赐教。

  1 #include <iostream>
  2 #include <string>
  3 
  4 #include <math.h>
  5 
  6 
  7 using namespace std;
  8 
  9 const char *afterDecimals[] = { "圆","角","分" };
 10 const char *chineseNum[] = {"零","壹","贰","叁","肆","伍","陆","柒","捌","玖"};
 11 const char *beforeDecimals[] = { "","拾","佰","仟"};
 12 const char *step[] = { "","万","亿","万","亿","万","亿" };
 13 
 14 bool isZero(long double num)
 15 {
 16     if (num > -0.0000001 && num < 0.0000001)
 17         return true;
 18     else
 19         return false;
 20 }
 21 
 22 int getLength(long double num)
 23 {
 24     int length = 0;
 25 
 26     //num零值判断
 27     if (isZero(num))
 28         return 1;
 29     else
 30     {
 31         /*
 32         while ((long long)num)
 33         {
 34             length++;
 35             num /= 10;
 36         }
 37         return length;
 38         */
 39         return (int)(log10(num) + 1);
 40     }
 41 }
 42 
 43 string change(long double num)
 44 {
 45     string str;
 46     char buf[64] = { 0 };
 47     sprintf_s(buf, "%0.0f", (num*100));
 48     int bufLength = strlen(buf);
 49     cout << "buf = " << buf << endl;
 50     int length = getLength(num);
 51     bool is0 = false;
 52 
 53     //小数点之前
 54     for (int i = 0; i < length; i++)
 55     {
 56         int j = buf[i] - '0';
 57         if (j == 0)
 58             is0 = true;
 59         else if (j > 0 && is0)//处理遇到连续多个0的情况
 60         {
 61             str += chineseNum[0] + string(chineseNum[j]) + beforeDecimals[(length - i - 1) % 4];
 62             is0 = false;
 63         }
 64         else
 65         {
 66             str += chineseNum[j];
 67             str += beforeDecimals[(length - i-1) % 4];
 68         }
 69         if ((length - i-1) % 4 == 0)
 70         {
 71             str += step[(length - i-1) / 4];
 72             is0 = false;
 73         }
 74     }
 75         
 76 
 77     //小数点以后
 78     for (int i = 0; i < bufLength; i++)
 79     {
 80         char ch = buf[i];
 81         if (i >= length)
 82             str += chineseNum[atoi(&ch)] + string(afterDecimals[i + 1 - length]);
 83         
 84         else if (i == length-1)
 85             str += string(afterDecimals[0]) + string(".");
 86         else
 87             continue;        
 88     }
 89     return str;
 90 }
 91 
 92 int main(int argc, char* argv[])
 93 {
 94     
 95     long double num = 123456789.236;
 96     cout << "num = 123456789.236" << endl;
 97     string str = change(num);
 98     cout << str << endl;
 99     
100 
101 
102     getchar();
103     return 0;
104 }

View Code

　　其中change(long double num)函数为主体，之所以采用long double类型是为了转换大于99999999的数字。