求两个任意长度的数乘积（还没有调试成功）

调了一个下午还没有调好，估计今天脑袋要晕掉，过几天接着调。想想思路上有啥好的改进没

// Console08172.cpp : Defines the entry point for the console application.
//
#include "stdafx.h"
#include <iostream>
#include <assert.h>
#include <math.h>
using namespace std;

const int N = 100;

void Calc2BigNum(int num1[], int len1, int num2[],int len2, int result[]);

int _tmain(int argc, _TCHAR* argv[])
{    
    int signed1 = 1;
    int signed2 = 1;
    //首先以字符形式存储
    char s1[N];
    char s2[N];
    //转换为整数型先存储
    int num1[N];
    int num2[N];
    //初始化数组
    for (int initnum = 0; initnum < N; ++initnum)
    {    
        s1[initnum] = '0';
        s2[initnum] = '0';
        num1[initnum] = -1;
        num2[initnum] = -1;
    }
    //结果存放数组
    int result[2*N] = {0};
    
    //记录小数点位置
    int dotloc1 = -1;
    int dotloc2 = -1;
    //记录小数点后的最终有效位置
    int dot1Afterloc = 0;
    int dot2Afterloc = 0;
    //记录小数点后的位数
    int dot1AfterNum = 0;
    int dot2AfterNum = 0;

    char ctemp;
    int i;
    
    cout << "请分别输入2个数(长度 <100),以#结束,若需要输入小数点，请以xxx.xxx的形式:\n" ;
    cout << "\n请输入第1个数：" ;
    cin >> ctemp;
    if (ctemp == '-')    { signed1 = -1;} 
    if (ctemp == '+')    { signed1 = 1; }
    else
    {
        assert(ctemp >='0' && ctemp <= '9');
        num1[0] = ctemp - '0';
        for (i = 1; i< N; ++i)
        {
            cin >> s1[i];
            if (s1[i] == '#') 
            {
                break;
            }
            if (s1[i] == '.')
            {
                dotloc1 = i;
                num1[i] = -2;
            }
            else
            {
                assert(s1[i] >= '0' && s1[i] <= '9');
                num1[i] = s1[i] - '0';
            }
        }
    }
    
    
    //确定小数点后面的第1个非0数的位置，
    if (dotloc1 != -1)
    {
        for (int cnt1 = N-1; num1[cnt1] != -2; --cnt1)    {dot1Afterloc = cnt1;}    
        dot1AfterNum = dot1Afterloc - dotloc1 + 1;
    }
    

    cout << "\n请输入第2个数：" ;
    int j;
    cin >> ctemp;
    if (ctemp == '-')    { signed2 = -1;} 
    if (ctemp == '+')    { signed2 = 1; }
    else
    {
        assert(ctemp >='0' && ctemp <= '9');
        num2[0] = ctemp - '0';
        for (j = 1; j< N; ++j)
        {
            cin >> s2[j];
            if (s2[j] == '#')
            {
                break;
            }
            if (s2[j] == '.')
            {
                dotloc2 = j;
                num2[j] = -2;
            }
            else
            {
                assert(s2[j] >= '0' && s2[j] <= '9');
                num2[j] = s2[j] - '0';
            }
        }
    }

    //小数点后面的第1个非0数的位置,如果有小数点的话
    if (dotloc2 != -1)
    {
        for (int cnt2 = N-1; num2[cnt2] != -2 && dotloc2 != -1; --cnt2)    {dot2Afterloc = cnt2;}
        dot2AfterNum = dot2Afterloc - dotloc2 + 1;

    }
    //小数点总数
    int dotsNum = dot1AfterNum + dot2AfterNum;
    //去小数点
    int k1 = 0;
    int k2 = 0;
    if (dotsNum != 0)
    {
        if (dotloc1 != -1)
        {
            for (k1 = dotloc1; k1 < N; ++k1)    { num1[k1] = num1[k1+1];}
            num1[k1] = -1;
        }
        if (dotloc2 != -1)
        {
            for (k2 = dotloc2; k2 < N; ++k2)    { num1[k2] = num2[k2+1];}    
            num2[k2] = -1;
        }
    }
    //数实际长度
    int len1 = 0;
    int len2 = 0;
    for (k1 = N-1; num1[k1] != -1; k1--) { ;} len1 = N-k1;
    for(k2 = N-1; num2[k2] != -1; k2--){;} len2 = N-k2;
    
    Calc2BigNum(num1, len1, num2, len2, result);

    //输出最终计算结果
    cout << "\n计算结果为：";
    if ((signed1*signed2) == -1)
    {
        cout << "-";
    }
    int locNot0 = 0;
    for (int rcnt = 0; rcnt< 2*N;rcnt++)
    {
        if (result[rcnt] != 0)
        {
            locNot0 = rcnt;
            break;
        }
    }
    for (int rcnt = locNot0; rcnt< 2*N;rcnt++)
    {
        if (rcnt == 2*N-1-dotsNum && dotsNum != 0)
        {
            cout << "." << result[rcnt];
        }
        else
            cout << result[rcnt];
    }
    cout << endl;

    return 0;
}

void Calc2BigNum( int num1[], int len1, int num2[],int len2, int result[] )
{
    int i;
    int j;
    int resultCnt = 2*N-1;
    int resultCtl = 2*N-1;
    int tmpNum = 0;

    for (i = len1-1; i >=0; i--)
    {
        resultCnt = resultCtl;
        for (j = len2-1; j >=0; j--)
        {
            tmpNum = num1[i]*num2[j];
            result[resultCnt] += tmpNum;
            if (tmpNum > 9)
            {
                result[resultCnt] = tmpNum%10;
                result[resultCnt-1] += tmpNum/10;
            }
            resultCnt--;
        }
        resultCtl--;
    }
}