每日一练——大数加减乘除运算实现（网易笔试题）

来源：互联网发布：磁力链接软件编辑：程序博客网时间：2024/05/30 19:32

前几天做网易笔试题时最后一道题是设计一个大数类，实现加减运算，因为做这道题时只有5分钟，结果我刚把类写出来，考试时间就结束了。其实在去年12月我写过一个RSA的加解密程序，其中就用到了大数的加减乘除运算（当然这只是RSA用到的一小部分），没有把最后一题写上去实在太可惜了(>﹏<。)～，今天把我设计的大数类贴出来，后面顺便附上面试时经常会让手写的加、减、乘代码。

一般的，我们会用一个很大的数组来表示一个大数，数组中的每一个数（0~9）来代表大数中的某一位，比如big_num[1000],那么这个数能够表示1000位大数，后来又人想到用数组中的每个数来表示0~99999999,这样数组只需要1000 / 8 = 125个元素即可实现。那么问题来了，一个unsigned long能表示的范围是0～4294967295也就是十六进制的0~0xFFFFFFFF,为什么我们不充分利用unsigned long能表示范围的的每一个数呢？

#define MAXLEN 64typedef unsigned char BYTE; class big_int{public:    ......    void set_zero();//大数清零    void modify_bit();//修正大数位数    big_int& mov(big_int &obj);//大数赋值大数    int cmp(big_int &obj);//两个大数进行比较，大于返回1，等于返回0，小于返回-1    big_int& add(big_int &num1, big_int &num2);//大数加大数 A=B+C    big_int& add(big_int &num);//大数加大数 A+=B    big_int& Sub(big_int &num1, big_int &num2);//大数减大数 A=B-C    big_int& Sub(big_int &num);//大数减大数 A-=B    big_int& mul(big_int &num1, big_int &num2);//大数乘大数 A=B*C    big_int& mul(big_int &num);//大数乘大数 A*=B    big_int& div(big_int &num1, big_int &num2);//大数除大数 A=B/C    big_int& div(big_int &num);//大数除大数 A/=B    ......            public:    int sign;//正数为1，负数为0    uint32_t data[MAXLEN];//数组中每一个数代表大数2^32次方进制的每一位    int length;//大数使用int的长度    int bit;//大数的二进制位位数};

//大数加大数A=B+Cbig_int& big_int::add(big_int &num1, big_int &num2){    uint32_t carry = 0;    uint64_t tmp;//用于存放两个无符号整形数(32位)的和，结果可能超过32位    set_zero();    length = num1.length > num2.length ? num1.length : num2.length;    for (int i = 0; i < length; i++)    {        tmp = (uint64_t)num1.data[i] + (uint64_t)num2.data[i] + (uint64_t)carry;        data[i] = (uint32_t)(tmp & 0x00000000FFFFFFFF);//data[i] = tmp % 0x100000000        carry = (uint32_t)(tmp >> 32);//carry = tmp / 0x100000000)    }    if (carry != 0)    {        data[length++] = carry;    }    modify_bit();    return *this;}//大数减大数 在减之前确保被减数大于减数A=B-Cbig_int& big_int::sub(big_int &num1, big_int &num2){    uint32_t carry = 0;    uint64_t tmp;    set_zero();    if (num1.cmp(num2) < 0)//如果被减数小于减数，交换两数再相减，符号位为负    {        big_int bi_tmp;        bi_tmp.mov(num1);        num1.mov(num2);        num2.mov(bi_tmp);        sign = -1;    }    int length = num1.length > num2.length ? num1.length : num2.length;    for (int i = 0; i < length; i++)    {        if ((uint64_t)num1.data[i] >= (uint64_t)num2.data[i] + (uint64_t)carry)        //如果obj.data[i] = 0xFFFFFFFF, carry = 1,相加会溢出，所以强转为uint64_t        {            data[i] = num1.data[i] - num2.data[i] - carry;            carry = 0;        }        else//minuend.data[i] < obj.data[i] + carry        {            tmp = num1.data[i] | 0x100000000;//tmp = num1.data[i] + 0x100000000;            tmp = tmp - num2.data[i] - carry;//tmp一定小于0x100000000            data[i] = (uint32_t)tmp;            carry = 1;        }    }    modify_bit();    return *this;}//两个大数相乘big_int& big_int::mul(big_int &num1, big_int &num2){    uint32_t carry;    uint64_t tmp;//两个32位二进制数相乘结果不会超过64位    big_int bi_tmp;    set_zero();    for (int i = 0; i < num2.length; i++)    {        carry = 0;        bi_tmp.set_zero();        for (int j = 0; j < num1.length; j++)        {            tmp = (uint64_t)num1.data[j] * (uint64_t)num2.data[i]            + (uint64_t)carry;            bi_tmp.data[bi_tmp.length++] = (uint32_t)(tmp & 0x00000000FFFFFFFF);            //等价于bi_tmp.data[bi_tmp.length++] = tmp % 0x100000000            carry = (uint32_t)(tmp >> 32);//等价于carry = tmp / 0x100000000)        }        if (carry != 0)        {        bi_tmp.data[bi_tmp.length++] = carry;        }        bi_tmp.left_move_len(i);        add(bi_tmp);    }    modify_bit();    return *this;}//大数除大数，采用试商的方法big_int& big_int::div(big_int &num1, big_int &num2){    big_int bi_tmp, bi_dividend;    int len;    uint64_t dividend_num;    uint32_t div_num;    bi_dividend.mov(num1);    set_zero();    while (bi_dividend.cmp(num2) > 0)    {        if (bi_dividend.data[bi_dividend.length-1] > num2.data[num2.length-1])        {        len = bi_dividend.length - num2.length;        //这里采用五入的试商方法        div_num = bi_dividend.data[bi_dividend.length-1] / (num2.data[num2.length-1]+1);        }        else if (bi_dividend.length > num2.length)        {            len = bi_dividend.length - num2.length - 1;            dividend_num = (uint64_t)bi_dividend.data[bi_dividend.length-1];            dividend_num = (dividend_num << 32) + bi_dividend.data[bi_dividend.length-2];            if (num2.data[num2.length-1] == 0xFFFFFFFF)            {            div_num = (uint32_t)(dividend_num >> 32);//dividend_num / 0x100000000 等价于 dividend_num >> 32            }            else            {            div_num = (uint32_t)(dividend_num / (uint64_t)(num2.data[num2.length-1]+1));            }        }        else//被除数最高位等于除数最高位 被除数位数等于除数位数        {            add(1);//被除数除以除数的商一定为1            break;        }        bi_tmp.mov(div_num);        bi_tmp.left_move_len(len);        add(bi_tmp);        bi_tmp.mul(num2);        bi_dividend.sub(bi_tmp);    }    if (bi_dividend.cmp(num2) == 0)    {        add(1);    }    modify_bit();    return *this;}

在面试和笔试中当然不能这么玩，附上面试时让手写的带符号的加减乘代码。

//无符号加法，且num1 >= num2string unsigned_add(string num1, string num2){    int len1 = num1.size();    int len2 = num2.size();            int carry = 0;    string result = "";    int tmp;    while (num1.size() != num2.size())//加0补齐    {        num2 = '0' + num2;    }    for (int i = len1 - 1; i >= 0; i--)    {        tmp = num1[i] - '0' + num2[i] - '0' + carry;        result += tmp % 10 + '0';        carry = tmp / 10;    }    if (carry)    {        result += carry + '0';    }    reverse(result.begin(), result.end());        return result;}//无符号减法，且num1 >= num2string unsigned_sub(string num1, string num2){    int len1 = num1.size();    int len2 = num2.size();            int carry = 0;    string result = "";    while (num1.size() != num2.size())//加0补齐    {        num2 = '0' + num2;    }    for (int i = len1 - 1; i >= 0; i--)    {        if (num1[i] >= num2[i] + carry)        {            result += num1[i] - num2[i] - carry + '0';            carry = 0;        }        else        {            result += 10 + num1[i] - num2[i] - carry + '0';            carry = 1;        }    }    while (result[result.size() - 1] == '0')//去掉多余的0    {        result = result.substr(0, result.size() - 1);    }    reverse(result.begin(), result.end());        return result;}//无符号乘法string unsigned_mul(string num1, string num2){    int len1 = num1.size();    int len2 = num2.size();    int cur, carry, tmp;    string result = "";    while (result.size() != len1 + len2)    {        result = '0' + result;    }    for (int i = len1 - 1; i >= 0; i--)    {        cur = len1 - 1 - i;        carry = 0;        for (int j = len2 - 1; j >= 0; j--)        {             tmp = result[cur] - '0' + (num1[i] - '0') * (num2[j] - '0') + carry;             result[cur] = tmp % 10 + '0';             carry = tmp / 10;             cur++;        }        if (carry)        {            result[cur++] = carry + '0';        }    }    while (result[result.size() - 1] == '0')//去掉多余的0    {        result = result.substr(0, result.size() - 1);    }    reverse(result.begin(), result.end());        return result;}//把符号位从string中分离开来int separate(string &num){    if (num[0] == '-')    {        num = num.substr(1, num.size());        return -1;    }    else    {        return 1;    }}//带符号的加法string add(string num1, string num2){    int sign1, sign2, tmp;    sign1 = separate(num1);    sign2 = separate(num2);    string result;    if (num1.size() < num2.size() || num1.size() == num2.size() && num1.compare(num2) < 0)    {        num1.swap(num2);        tmp = sign1;        sign1 = sign2;        sign2 = tmp;    }    if (sign1 == -1 && sign2 == 1)    {        result = '-' + unsigned_sub(num1, num2);    }    else if (sign1 == 1 && sign2 == -1)    {        result = unsigned_sub(num1, num2);    }    else if (sign1 == 1 && sign2 == 1)    {        result = unsigned_add(num1, num2);    }    else    {        result = '-' + unsigned_add(num1, num2);    }    return result;}//带符号的减法string sub(string num1, string num2){    int sign1, sign2, tmp;    sign1 = separate(num1);    sign2 = separate(num2);    string result;    if (num1.size() < num2.size() || num1.size() == num2.size() && num1.compare(num2) < 0)    {        num1.swap(num2);        tmp = sign1;        sign1 = sign2 * (-1);        sign2 = tmp * (-1);    }    if (sign1 == -1 && sign2 == 1)    {        result = '-' + unsigned_add(num1, num2);    }    else if (sign1 == 1 && sign2 == -1)    {        result = unsigned_add(num1, num2);    }    else if (sign1 == 1 && sign2 == 1)    {        result = unsigned_sub(num1, num2);    }    else    {        result = '-' + unsigned_sub(num1, num2);    }    return result;}//带符号的乘法string mul(string num1, string num2){    int sign1, sign2, tmp;    sign1 = separate(num1);    sign2 = separate(num2);    string result;    if (num1.size() < num2.size() || num1.size() == num2.size() && num1.compare(num2) < 0)    {        num1.swap(num2);        tmp = sign1;        sign1 = sign2;        sign2 = tmp;    }    result = unsigned_mul(num1, num2);        if (sign1 * sign2 == -1)    {        result = '-' + result;    }    return result;}

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