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。

#include <iostream>
#include <string>

#include <math.h>


using namespace std;

const char *afterDecimals[] = { "圆","角","分" };
const char *chineseNum[] = {"零","壹","贰","叁","肆","伍","陆","柒","捌","玖"};
const char *beforeDecimals[] = { "","拾","佰","仟"};
const char *step[] = { "","万","亿","万","亿","万","亿" };

bool isZero(long double num)
{
    if (num > -0.0000001 && num < 0.0000001)
        return true;
    else
        return false;
}

int getLength(long double num)
{
    int length = 0;

    //num零值判断
    if (isZero(num))
        return 1;
    else
    {
        /*
        while ((long long)num)
        {
            length++;
            num /= 10;
        }
        return length;
        */
        return (int)(log10(num) + 1);
    }
}

string change(long double num)
{
    string str;
    char buf[64] = { 0 };
    sprintf_s(buf, "%0.0f", (num*100));
    int bufLength = strlen(buf);
    cout << "buf = " << buf << endl;
    int length = getLength(num);
    bool is0 = false;

    //小数点之前
    for (int i = 0; i < length; i++)
    {
        int j = buf[i] - '0';
        if (j == 0)
            is0 = true;
        else if (j > 0 && is0)//处理遇到连续多个0的情况
        {
            str += chineseNum[0] + string(chineseNum[j]) + beforeDecimals[(length - i - 1) % 4];
            is0 = false;
        }
        else
        {
            str += chineseNum[j];
            str += beforeDecimals[(length - i-1) % 4];
        }
        if ((length - i-1) % 4 == 0)
        {
            str += step[(length - i-1) / 4];
            is0 = false;
        }
    }
        

    //小数点以后
    for (int i = 0; i < bufLength; i++)
    {
        char ch = buf[i];
        if (i >= length)
            str += chineseNum[atoi(&ch)] + string(afterDecimals[i + 1 - length]);
        
        else if (i == length-1)
            str += string(afterDecimals[0]) + string(".");
        else
            continue;        
    }
    return str;
}

int main(int argc, char* argv[])
{
    
    long double num = 123456789.236;
    cout << "num = 123456789.236" << endl;
    string str = change(num);
    cout << str << endl;
    


    getchar();
    return 0;
}

　　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。