书本p119相关题目，但我由于时间关系，没有考虑输入的数字中有负数的情况。

思路：

1. (用字符串存储大数)将字符串number1与number2对齐，在长度较小的字符串上高位补0，生成新的两个长度一致的字符串；

2. 字符串从低位相加，产生每位的相加结果及其进位，最后判断是否在最高位有进位。
   代码：

   void addBigDigit(char* number1, char* number2, int length1, int length2);
   void printChar(char* number);
   //两个字符串表示的大数相加
   void addBigDigit(char* number1, char* number2, int length1, int length2)
   {
    int maxLength = length1;
    if (length1 < length2)
        maxLength = length2;
   
    //新建两个字符串存储补齐0之后长度一致的字符串
    char* fullNumber1 = new char[maxLength + 1];
    char* fullNumber2 = new char[maxLength + 1];
    //初始化字符串，这个很重要，因为后续将number1,number2 copy到新字符串中时会覆盖掉这里初始化的0，就不用了特意往高位补0了。
    memset(fullNumber1, '0', maxLength);
    fullNumber1[maxLength] = '\0';
    memset(fullNumber2, '0', maxLength);
    fullNumber2[maxLength] = '\0';
   
    //复制旧字符串到新字符串中
    if (maxLength != length1)
    {
        int i = maxLength - 1, j = length1 - 1;
        while (i>=0&&j>=0)
        {
            fullNumber1[i--] = number1[j--];
        }
        strcpy(fullNumber2, number2);
    }
   
    if (maxLength != length2)
    {
        int i = maxLength - 1, j = length2 - 1;
        while (i >= 0 && j >= 0)
        {
            fullNumber2[i--] = number2[j--];
        }
        strcpy(fullNumber1, number1);
    }
    //刚开始忘了这个，若没有，则新字符串就全是0了
    if(maxLength == length2&&maxLength==length1)
    {
        strcpy(fullNumber2, number2);
        strcpy(fullNumber1, number1);
    }
   
    //printChar(fullNumber1);
    //printChar(fullNumber2);
   
    //本位相加结果和进位结果
    int nSum = 0,nTakeOver = 0;
    //判断在最高位是否越界，即是否产生了新的超过字符串长度的最高位
    bool isOverflow = false;
    //相加后的结果字符串
    char* addNumber = new char[maxLength+1];
    memset(addNumber, '0', maxLength);
    addNumber[maxLength] = '\0';
   
    for (int i = maxLength - 1; i >= 0; i--)
    {
   
        nSum = fullNumber1[i] - '0' + fullNumber2[i] - '0' + nTakeOver;
        //相加完之后设置进位为0
        nTakeOver = 0;
        if (nSum >= 10)
        {
            nSum -= 10;
            nTakeOver = 1;
        }
        addNumber[i] = nSum + '0';
        //溢出：计算到最高位且进位不为0
        if (i == 0 && nTakeOver > 0)
            isOverflow = true;
    }
   
    if (isOverflow)
        cout << '1';
    printChar(addNumber);
    return;
   }
   void printChar(char* number)
   {
    for (int i = 0; i < strlen(number); i++)
        cout << number[i];
    cout << endl;
   }
   int main()
   {
    char number1[] = "1234567";
    char number2[] = "46781";
    int length1 = strlen(number1);
    int length2 = strlen(number2);
    addBigDigit(number1, number2, length1,length2);
   }

更新后，用数组表示大数

/*
大数相加问题:
1.大数表示：数组表示
2.大数相加：较少位数高位补0；相加进位；是否溢出；
*/

//高位补0
void CompleteZero(vector<int> &number,int len,int bigLen)
{
    vector<int> completeNumber(bigLen, 0);
    int gap = bigLen - len;
    for (int i = 0; i < len; ++i)
    {
        completeNumber[i + gap] = number[i];
    }
    number.clear();
    number.reserve(bigLen);
    number = completeNumber;
}

//大数相加
void AddBigData(vector<int> number1, vector<int> number2)
{
    int len1 = number1.size(), len2 = number2.size();
    if (len1 < len2)
    {
        CompleteZero(number1, len1, len2);
    }

    if (len1 > len2)
        CompleteZero(number2, len2,len1);

    int len = max(len1, len2);
    //存储相加结果,多一位是怕有溢出
    vector<int> addResult(len + 1, 0);

    //判断最高位是否会溢出
    bool isOverFlow = false;
    //本位对下一位的进位
    int nTakeOver = 0;
    for (int i = len-1; i >= 0; --i)
    {
        //本位相加和，可能会大于等于10
        int nSum = number1[i] + number2[i]+nTakeOver;
        //cout << nSum << "   " << nTakeOver << endl;
        if (nSum >= 10)
        {
            nTakeOver = 1;
            nSum -= 10;
            //若是在最高位有进位，则溢出
            if (i == 0)
            {
                isOverFlow = true;
            }
            addResult[i+1] = nSum;
        }
        else
        {
            nTakeOver = 0;
            addResult[i+1] = nSum;
        }
    }

    int i=1;
    if (isOverFlow)
        i = 0;
    for (; i < len + 1; ++i)
        cout << addResult[i];
    cout << endl;
}

int main()
{
    vector<int> number1 = { 1,2,3,4,5,6,7};
    vector<int> number2 = { 5,6,7,8 };
    AddBigData(number1, number2);
    return 0;
}