实现大整数相加（考虑符号位，可能有负整数） 思维严谨程度！！

//实现大整数相加
//还得考虑符号位，一个比另一个短
#include<iostream>
#include <string>
#include <cstring>

using namespace std;
#define maxlen 2001
int a[maxlen];
int b[maxlen];
int len1, len2, i, j;

int bigger(int a, int b)
{
    return a>b ? a : b;
}

void Add(int underzero)
{
    if (underzero)
        cout << '-';
    int big = bigger(len1, len2);
    for (i = 0; i <= big; i++)
    {
        a[i] += b[i];
        if (a[i] >= 10)
        {
            a[i] -= 10;
            a[i + 1]++;
        }
    }

    bool flag = false;//用来忽略高位的0，中间的0才是有效数字
    for (i = big; i >= 0; i--)
    {
        if (flag)
            cout << a[i];
        else if (a[i])
        {
            cout << a[i];
            flag = true;
        }
    }
    cout << endl;
}

void Minus(int aa[], int bb[], int biglen, int underzero)
{
    if (underzero)
        cout << '-';
    for (i = 0; i <= biglen; i++)
    {
        if (aa[i] >= bb[i])
            aa[i] -= bb[i];
        else
        {
            aa[i] = aa[i] + 10 - bb[i];
            aa[i + 1]--;
        }
    }

    bool flag = false;//同样去掉最高位的无效‘1’
    for (i = biglen; i >= 0; i--)
    {
        if (flag)
            cout << aa[i];
        else if (aa[i])
        {
            cout << aa[i];
            flag = true;
        }
    }
    cout << endl;
}

int main()
{
    string s1, s2;
    bool success;
    while (cin >> s1 >> s2)
    {
        memset(a, 0, sizeof(a));
        memset(b, 0, sizeof(b));
        len1 = s1.length();
        len2 = s2.length();

        for (i = len1 - 1, j = 0; i>0; i--)
            a[j++] = s1[i] - '0';
        if (s1[0] != '-')
            a[j] = s1[0] - '0';
        for (i = len2 - 1, j = 0; i>0; i--)
            b[j++] = s2[i] - '0';
        if (s2[0] != '-')
            b[j] = s2[0] - '0';

        if (s1[0] == '-' && s2[0] == '-')
            Add(1);  //和为负
        else if (s1[0] != '-' && s2[0] != '-')
            Add(0);  //和为正
        else if (s1[0] == '-')
        {
            if (len1 - 1 > len2) //负数大
                Minus(a, b, len1 - 1, 1);
            else if (len1 - 1 < len2) //正数大
                Minus(b, a, len2, 0);
            else
            {
                success = true;
                for (i = len1 - 2, j = len2 - 1; j >= 0; j--, i--) //这里的len-2是个关键，len-1是长度，之所以再减去1，是要减去负号的位置
                {
                    if (a[i]>b[j]) //负数大
                    {
                        success = false;
                        Minus(a, b, len2, 1);
                        break;
                    }
                    else if (a[i]<b[j]) //正数大
                    {
                        success = false;
                        Minus(b, a, len2, 0);
                        break;
                    }
                }
                if (success) //相等
                    cout << "0" << endl;
            }
        }
        else if (s2[0] == '-')
        {
            if (len2 - 1 > len1) //负数大
                Minus(b, a, len2 - 1, 1);
            else if (len2 - 1 < len1) //正数大
                Minus(a, b, len1, 0);
            else
            {
                success = true;
                for (i = len2 - 2, j = len1 - 1; j >= 0; j--, i--) //这里的len-2是个关键，len-1是长度，之所以再减去1，是要减去负号的位置
                {
                    if (a[j]>b[i]) //正数大
                    {
                        success = false;
                        Minus(a, b, len1, 0);
                        break;
                    }
                    else if (a[j]<b[i]) //负数大
                    {
                        success = false;
                        Minus(b, a, len1, 1);
                        break;
                    }
                }
                if (success) //相等
                    cout << "0" << endl;
            }
        }
    }
    return 0;
}

//实现大整数相加
//还得考虑符号位，一个比另一个短
#include<iostream>
#include <string>
#include <cstring>

using namespace std;
#define maxlen 2001
int a[maxlen];
int b[maxlen];
int len1, len2, i, j;

int bigger(int a, int b)
{
    return a>b ? a : b;
}

void Add(int underzero)
{
    if (underzero)
        cout << '-';
    int big = bigger(len1, len2);
    for (i = 0; i <= big; i++)
    {
        a[i] += b[i];
        if (a[i] >= 10)
        {
            a[i] -= 10;
            a[i + 1]++;
        }
    }

    bool flag = false;//用来忽略高位的0，中间的0才是有效数字
    for (i = big; i >= 0; i--)
    {
        if (flag)
            cout << a[i];
        else if (a[i])
        {
            cout << a[i];
            flag = true;
        }
    }
    cout << endl;
}

void Minus(int aa[], int bb[], int biglen, int underzero)//数组的元素下标从小到大对应大整数（低位->高位）哪一个数的绝对值大，那个数就放前面，underzero=1表示最终结果为负数
{
    if (underzero)
        cout << '-';
    for (i = 0; i <= biglen; i++)
    {
        if (aa[i] >= bb[i])
            aa[i] -= bb[i];
        else
        {
            aa[i] = aa[i] + 10 - bb[i];
            aa[i + 1]--;
        }
    }

    bool flag = false;//同样去掉最高位的无效‘1’
    for (i = biglen; i >= 0; i--)
    {
        if (flag)
            cout << aa[i];
        else if (aa[i])
        {
            cout << aa[i];
            flag = true;
        }
    }
    cout << endl;
}

int main()
{
    string s1, s2;
    bool success;
    while (cin >> s1 >> s2)
    {
        memset(a, 0, sizeof(a));
        memset(b, 0, sizeof(b));
        len1 = s1.length();
        len2 = s2.length();

        for (i = len1 - 1, j = 0; i>0; i--)//将大整数的最低位（个位）存入到数组下标最小的位置
            a[j++] = s1[i] - '0';
        if (s1[0] != '-')
            a[j] = s1[0] - '0';
        for (i = len2 - 1, j = 0; i>0; i--)
            b[j++] = s2[i] - '0';
        if (s2[0] != '-')
            b[j] = s2[0] - '0';

        if (s1[0] == '-' && s2[0] == '-')
            Add(1);  //和为负
        else if (s1[0] != '-' && s2[0] != '-')
            Add(0);  //和为正
        else if (s1[0] == '-')
        {
            if (len1 - 1 > len2) //负数大
                Minus(a, b, len1 - 1, 1);
            else if (len1 - 1 < len2) //正数大
                Minus(b, a, len2, 0);
            else
            {
                success = true;
                for (i = len1 - 2, j = len2 - 1; j >= 0; j--, i--) //这里的len-2是个关键，len-1是长度，之所以再减去1，是要减去负号的位置
                {
                    if (a[i]>b[j]) //负数大
                    {
                        success = false;
                        Minus(a, b, len2, 1);
                        break;
                    }
                    else if (a[i]<b[j]) //正数大
                    {
                        success = false;
                        Minus(b, a, len2, 0);
                        break;
                    }
                }
                if (success) //相等
                    cout << "0" << endl;
            }
        }
        else if (s2[0] == '-')
        {
            if (len2 - 1 > len1) //负数大
                Minus(b, a, len2 - 1, 1);
            else if (len2 - 1 < len1) //正数大
                Minus(a, b, len1, 0);
            else
            {
                success = true;
                for (i = len2 - 2, j = len1 - 1; j >= 0; j--, i--) //这里的len-2是个关键，len-1是长度，之所以再减去1，是要减去负号的位置
                {
                    if (a[j]>b[i]) //正数大
                    {
                        success = false;
                        Minus(a, b, len1, 0);
                        break;
                    }
                    else if (a[j]<b[i]) //负数大
                    {
                        success = false;
                        Minus(b, a, len1, 1);
                        break;
                    }
                }
                if (success) //相等
                    cout << "0" << endl;
            }
        }
    }
    return 0;
}