超长浮点数大数相乘

代码：

#include 
#include 
#include 

using namespace std;

static string BigMulti(string &a, string &b)
{
    string ret;
    vector<int> v(a.length() + b.length() - 1, 0);
    for (int i = 0; i < a.length(); i++) {
        for (int j = 0; j < b.length(); j++) {
            v[i + j] += (a[i] - '0') * (b[j] - '0');
        }
    }

    int upper = 0;
    int tmp;
    char c;
    for (int i = v.size() - 1; i >= 0; i--) {
        tmp = upper + v[i];
        c = (tmp % 10 + '0');
        ret = c + ret;
        upper = tmp / 10;
    }

    if (upper > 0) {
        ret = to_string(upper) + ret;
    }
    return ret;
}

int main()
{
    string str;
    cin >> str;
    string a, b;
    // 用于统计小数点位数
    int cntPoint = 0;
    // 第一个乘数小数点距离末尾位数
    int cntFirstVarPoint;
    // 第二个乘数小数点距离末尾位数
    int cntSecondVarPoint;

    // 分割乘数为a和b
    size_t idx = str.find_first_of('*');
    a = str.substr(0, (idx - 1) - 0 + 1);
    b = str.substr(idx + 1, (str.size() - 1) - (idx + 1) + 1);
    int aLength = a.size();
    int bLength = b.size();

    //1.找出2个乘数的小数点位数   2.把a和b转换成两个整数string
    size_t idxPoint1 = a.find_first_of('.');
    size_t idxPoint2 = b.find_first_of('.');
    if (idxPoint1 == string::npos) {
       cntFirstVarPoint = 0;
    } else {
        cntFirstVarPoint = aLength - 1 - idxPoint1;
        a.erase(idxPoint1, 1);
    }
    if (idxPoint2 == string::npos) {
       cntSecondVarPoint = 0;
    } else {
        cntSecondVarPoint = bLength - 1 - idxPoint2;
        b.erase(idxPoint2, 1);
    }
    cntPoint = cntFirstVarPoint + cntSecondVarPoint;
    int j = cntPoint;

    string result = BigMulti(a, b);

    //计算小数点后0的个数
    int i = result.size() - 1;
    int cntZero = 0;
    while (j--) {
        if (result[i] == '0') {
            cntZero++;
            i--;
        } else {
            break;
        }
    }

    if (cntZero == cntPoint) {
        result.erase(result.size() - cntZero, cntZero);
    } else {
        result.insert(result.size() - cntPoint, 1, '.');
        result.erase(result.size() - cntZero, cntZero);
    }
    cout << result << endl;
    return 0;
}