C++金额的中文大写

学习、记录、成长。

这道题还是做了好半天的，踩了很多细节问题...以前拿C写的时候好像用函数为了存大写数字就整的乱七八糟，这次本以为用string会大有增进，不成想在各种修补下依旧变得乱七八糟。但我觉得框架还是很清晰（拍桌）看代码就懂。

#include <iostream>
#include <string>
using namespace std;
int main()
{
    string big[] = {"整", "元", "拾", "佰", "仟", "万", "拾", "佰", "仟", "亿", "拾", "佰", "仟", "万"};//整数单位
    string money[] = {"零", "壹", "贰", "叁", "肆", "伍", "陆", "柒", "捌", "玖"};//大写数字
    string small[] = {"角", "分"};//小数单位
    string data;  //保存输入数据
    string ans[2000];  //输出缓存
    int res, big_len, small_len, num, fx, f0;//整数位长、小数位长、金额转换、小数位标记、零标记
    char numm;  //转换用中介
    while (cin >> data)
    {
        fx = 0;
        f0 = 1;
        res = data.find(".");

        if (res > 0)   //这个if-else 判断是否有小数位
        {
            small_len = data.length() - 2 - res;
            fx = 1;
            big_len = res;
        }
        else
            big_len = data.length();

        int i, j, k1 = 0, k2 = 0;  //分别用来遍历原字符串与答案字符串

        for (i = 0; i < big_len; i++)   //我是一个小裁缝、缝补本领强。。  
        {
            numm = data[i];
            num = numm - '0';
            if (num == 0 && i == 0)
            {
                ans[0] = "零";
                ans[1] = "元";
                ans[2] = "整";
                k1 = 2;
                //cout << "零元整" << endl;
                break;
            }
            if (num == 0)
            {
                if (k1 >= 2 && ans[k1 - 2] == "壹" && ans[k1 - 1] == "拾")
                {
                    ans[k1 - 2] = ans[k1 - 1];
                    k1--;
                    if ((big_len - i) == 1 || (big_len - i) == 5 || (big_len - i) == 9)
                        ans[k1++] = big[big_len - i];
                    continue;
                }
                if ((big_len - i) == 1 || (big_len - i) == 5 || (big_len - i) == 9)
                    ans[k1++] = big[big_len - i];
                if (f0 == 1)
                {
                    f0 = 0;
                }
            }
            else if (num > 0)
            {
                if (f0 == 0)
                {
                    ans[k1++] = money[0];
                    f0 = 1;
                }

                ans[k1++] = money[num];
                ans[k1++] = big[big_len - i];
            }
        }

        if (fx)
        {
            i++;
            for (j = 0; j <= small_len; i++, j++)
            {
                numm = data[i];
                num = numm - '0';
                if (num == 0)
                {
                    if (f0 == 1)
                    {
                        f0 = 0;
                    }
                }
                else if (num > 0)
                {
                    if (f0 == 0)
                    {
                        ans[k1++] = money[0];
                        ans[k1++] = small[j-1];
                        f0 = 1;
                    }
                    ans[k1++] = money[num];
                    ans[k1++] = small[j];
                }
            }
        }
        else
            ans[k1++] = big[0];
        for (k2 = 0; k2 < k1; k2++)
            cout << ans[k2];
        cout << endl;
    }
    //system("pause");
    return 0;
}