100位有符号整形数加法（数组）

//加法器，若要变为减法器，只需改动相应的判断条件即可
#include <iostream>
using namespace std;
struct num//储存多位的整形数字
{
    char input[102];
    char sign;
    char data[101];
    int len;
};
struct num plus_(struct num a, struct num b);//无符号整形的加法
struct num minus_(struct num a, struct num b);//无符号整形的减法，要求a>b
int main()
{
    while (1)//while循环用作多次测试
    {
        struct num a, b, c;//a,b为加数,c储存二者相加结果
        cin >> a.input;
        cin >> b.input;
        if (a.input[0] == '-')//把a和b的符号存起来，把去掉符号的数放到结构体的另一个数组里面
        {
            a.sign = '-';
            for (int i = 1; i <= strlen(a.input); i++)
            {
                a.data[i - 1] = a.input[i];
            }
        }
        else
        {
            a.sign = '+';
            for (int i = 0; i <= strlen(a.input); i++)
            {
                a.data[i] = a.input[i];
            }
        }
        if (b.input[0] == '-')
        {
            b.sign = '-';
            for (int i = 1; i <= strlen(b.input); i++)
            {
                b.data[i - 1] = b.input[i];
            }
        }
        else
        {
            b.sign = '+';
            for (int i = 0; i <= strlen(b.input); i++)
            {
                b.data[i] = b.input[i];
            }
        }
        //根据a,b符号的各种情况，调用plus_e和minus_函数进行运算
        if (a.sign == '+' && b.sign == '+')
        {//a,b为正数，直接相加
            c = plus_(a, b);
            const int len_ = strlen(c.data);
            int i = 0;
            for (i = 0; i < len_; i++)
            {
                if ((len_ - i) % 4 == 0 && i!=0)
                    cout << ',';
                cout << c.data[i];
            }
            cout << endl << "位数：" << strlen(a.data) << "，" << strlen(b.data) << "，" << i;
        }
        else if (a.sign == '-' && b.sign == '-')
        {//a,b皆为负数，则a,b去掉负号后相加，在结果前添一个负号
            c = plus_(a, b);
            if (c.data[0] != 0)
                cout << '-';
            int i = 0;
            const int len_ = strlen(c.data);
            for (i = 0; i < len_; i++)
            {
                if ((len_ - i) % 4 == 0 && i != 0)
                    cout << ',';
                cout << c.data[i];
            }
            cout << endl << "位数：" << strlen(a.data) << "，" << strlen(b.data) << "，" << i;
        }
        else
        {//a,b一正一负,将a,b去掉负号后，用它们之间大的减去小的
            int e_flag = 0;//a比b长（也即a>b)则为1，a比b短（即a<b)则为-1
            int z_flag = 0;//结果为0标志
            int la = strlen(a.data);
            int lb = strlen(b.data);
            if (la > lb)e_flag = 1;
            else if (la < lb)e_flag = -1;
            else//若a,b等长
            {//下面判断它们的大小
                for (int i = 0, j = 0; i < la, j < lb; i++, j++)
                {
                    if (a.data[i] > b.data[j])
                    {
                        e_flag = 1;//a>b
                        break;
                    }
                    else if (a.data[i] < b.data[j])
                    {
                        e_flag = -1;//a<b
                        break;
                    }
                }
            }
            if (e_flag == 1)//a>b,则c=a-b
            {
                c = minus_(a, b);
                if (a.sign == '-' && b.sign == '+')//如果a负b正且a大于b，输出'-'
                    cout << '-';
            }
            else if (e_flag == -1)//a<b，则c=b-a,结果前面添负号
            {
                c = minus_(b, a);
                if (a.sign == '+' && b.sign == '-')//如果a正b负且b大于a，输出'-'
                    cout << '-';
            }
            else//a,b相等的情况
            {
                c.data[0] = '0';
                c.data[1] = '';
            }
            int k = -1;//k为数组下标
            while (c.data[++k] == '0')//k移动到第一个不为'0'的地方
            {
                if (k == strlen(c.data) - 1)//如果k移到了数组尾部，说明全为0
                {
                    z_flag = 1;// ”为零“标志置1
                    break;
                }
            }
            int cnt = 0;//输出计数器，代表结果的位数，也用于控制','的输出
            const int len_ = strlen(c.data);
            for (int i = k; i < len_; i++)
            {
                if ((len_ - i) % 4 == 0 && i!=k)
                    cout << ',';
                cout << c.data[i];
                cnt++;
            }
            cout << endl << "位数：" << strlen(a.data) << "，" << strlen(b.data) << "，";
            cout << cnt;
        }
        cout << endl << endl;
    }
    return 0;
}
struct  num plus_(struct num a, struct num  b)
{
    int la = strlen(a.data);
    int lb = strlen(b.data);
    struct num c = {};
    char s[103];//存放计算结果的栈
    int r = -1;//栈顶指针
    int cx = 0;//进位标志，须初始化为0
    int i, j;
    for (i = la - 1, j = lb - 1; i >= 0 && j >= 0; i--, j--)
    {
        int t1 = a.data[i] - '0';// -'0'操作将char转换为int
        int t2 = b.data[j] - '0';
        int t3 = t1 + t2;
        if (t3 + cx >= 10)//如果a，b同一位相加大于等于10，则相加结果-10并将cx置1
        {
            t3 = t3 + cx - 10;
            cx = 1;
        }
        else//否则将cx置0
        {
            cx = 0;
        }
        s[++r] = t3 + '0';//将此位相加结果压入s
    }
    if (i < 0)//如果b比a长，则将b剩余数据放入s
    {
        for (int k = j; k >= 0; k--)
        {
            int t = b.data[k] - '0';
            if (t + cx >= 10)//注意遗留的cx
            {
                t = t + cx - 10;
                cx = 1;
            }
            else
            {
                t = t + cx;
                cx = 0;
            }
            s[++r] = t + '0';
        }
    }
    else if (j < 0)//同上
    {
        for (int k = i; k >= 0; k--)
        {
            int t = a.data[k] - '0';
            if (t + cx >= 10)
            {
                t = t + cx - 10;
                cx = 1;
            }
            else
            {
                t = t + cx;
                cx = 0;
            }
            s[++r] = t + '0';
        }
    }
    if (cx == 1)//第一位数（最高位）是不是有进位，如果是则压入s
    {
        s[++r] = cx + '0';
    }
    int i1, i2;//将栈s的数据放到结果c的data数组中
    for (i1 = 0, i2 = r; i2 >= 0; i1++, i2--)
    {
        c.data[i1] = s[i2];
    }
    c.data[i1] = '';
    return c;
}
struct num minus_(struct num a, struct num b)//与plus_函数相似，+变为-，cx由进位变借位
{
    int la = strlen(a.data);
    int lb = strlen(b.data);
    struct num c = {};
    char s[103];
    int r = -1;
    int cx = 0;
    int i, j;
    for (i = la - 1, j = lb - 1; i >= 0 && j >= 0; i--, j--)
    {
        int t1 = a.data[i] - '0';
        int t2 = b.data[j] - '0';
        int t3 = t1 - t2;
        if (t3 - cx < 0)
        {
            t3 = t3 - cx + 10;
            cx = 1;
        }
        else
        {
            t3 = t3 - cx;
            cx = 0;
        }
        s[++r] = t3 + '0';
    }
    if (i < 0)
    {
        for (int k = j; k >= 0; k--)
        {
            int t = b.data[k] - '0';
            if (t - cx < 0)
            {
                t = t - cx + 10;
                cx = 1;
            }
            else
            {
                t = t - cx;
                cx = 0;
            }
            s[++r] = t + '0';
        }
    }
    else if (j < 0)
    {
        for (int k = i; k >= 0; k--)
        {
            int t = a.data[k] - '0';
            if (t - cx < 0)
            {
                t = t - cx + 10;
                cx = 1;
            }
            else
            {
                t = t - cx;
                cx = 0;
            }
            s[++r] = t + '0';
        }
    }
    int i1, i2;
    for (i1 = 0, i2 = r; i2 >= 0; i1++, i2--)
    {
        c.data[i1] = s[i2];
    }
    c.data[i1] = '';
    return c;
}