十进制转换为N进制问题

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问题描述：给定十进制数(非负的实数)的字符串表示与转换的进制，转换的精度，要求输出转换后的进制数的字符表示

代码：

char unsignedToChar(unsigned n){    char ch = '#';    if (0 <= n && n <= 9)        ch = '0' + n;    else    if (10 <= n && n <= 36)        ch = n - 10 + 'A';    return ch;}unsigned charToUnsigned(char ch){    unsigned n = 0;    if ('0' <= ch && ch <= '9')        n = ch - '0';    else    if ('A' <= ch && ch <= 'Z')        n = ch - 'A' + 10;    return n;}std::string decimalTo(const std::string &decimal, unsigned digit, unsigned accuracy) {    unsigned accuracyCopy = accuracy;    std::stringstream strToDecimal(decimal);    std::stack<char> bInt;    std::string decNumStr;    unsigned integer;    double decNum, decimalPlace;    const unsigned digitNum = digit;    /* 字符串流将十进制字符串表示转换为double类型 */    strToDecimal >> decNum;    /* 获取十进制数的整数与小数部分 */    integer = static_cast<unsigned>(decNum / 1);    decimalPlace = decNum - integer;    /* 十进制整数部分转换为二进制 */    while (integer)    {        bInt.push(unsignedToChar(integer % digitNum));        integer /= digitNum;    }    while (!bInt.empty())    {        decNumStr.push_back(bInt.top());        bInt.pop();    }    /* 十进制小数部分转换为二进制 */    if (accuracyCopy != 0)        decNumStr.push_back('.');    while (decimalPlace && accuracyCopy != 0)    {        --accuracyCopy;        decimalPlace *= digitNum;        decNumStr.push_back(unsignedToChar(static_cast<char>(decimalPlace / 1)));        decimalPlace -= static_cast<int>(decimalPlace) / 1;    }    while (accuracyCopy != 0)    {        --accuracyCopy;        decNumStr.push_back(unsignedToChar(0));    }    return decNumStr;}</span>

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