大数的运算

//模拟加法的运算
string add(string num1, string num2)
{
    //位数补齐
    int len1 = num1.size();
    int len2 = num2.size();
    int diffNum = abs(len1 - len2);
    int longSize = len1 > len2 ? len1 : len2;
    if (len1 < len2)
    {
        num1.insert(0, diffNum, ‘0‘);
    }
    else if (len2 < len1)
    {
        num1.insert(0, diffNum, ‘0‘);
    }
    string ret;
    ret.resize(longSize);
    //逐位相加
    int step = 0;
    for (int i = longSize - 1; i >= 0; i--)
    {
        ret[i] = (num1[i] - ‘0‘) + (num2[i] - ‘0‘) + step;
        ret[i] += ‘0‘;
        //更新
        if (ret[i] > ‘9‘)
        {
            ret[i] -= 10;
            step = 1;
        }
        else step = 0;
    }
    if (step == 1)
    {
        ret.insert(0, 1, ‘1‘);
    }
    return ret;
}

 1 //模拟减法操作
 2 string sub(string num1, string num2)
 3 {
 4     //位数补齐
 5     int len1 = num1.size();
 6     int len2 = num2.size();
 7     int diffNum = abs(len1 - len2);
 8     int longSize = len1 > len2 ? len1 : len2;
 9     if (len1 < len2)
10     {
11         num1.insert(0, diffNum, ‘0‘);
12     }
13     else if (len2 < len1)
14     {
15         num1.insert(0, diffNum, ‘0‘);
16     }
17     string ret;
18     ret.resize(longSize);
19     //逐位相减
20     for (int i = longSize - 1; i >= 0; i--)
21     {
22         //判断是否需要借位
23         if (num1[i] < num2[i])
24         {
25             num1[i] += 10;
26             //更新高位
27             num1[i - 1]--;
28         }
29         ret[i] = (num1[i] - ‘0‘) - (num2[i] - ‘0‘);//相减完后变回字符
30 
31     }
32     //删除前置0
33     while (ret.size() > 1 && ret[0] == ‘0‘)
34     {
35         ret.erase(0, 1);
36     }
37     return ret;
38 }

 1 pair<string, string> dev(string num1, string num2)
 2 {
 3     string ret; //商
 4     string rem = num1; //余数
 5     int diffNum = num1.size() - num2.size();//给除数放大
 6     num2.append(diffNum, ‘0‘);
 7     for (int i = 0; i <= diffNum; i++)
 8     {
 9         //记录减法执行的次数
10         char count = ‘0‘;
11         while (1)
12         {
13             if (less(rem, num2))
14             {
15                 break;
16             }
17             rem = sub(rem, num2);
18             count++;
19         }
20         ret += count;
21         num2.pop_back();//除数减小10倍
22     }
23     //去零
24     while (ret.size() > 1 && ret[0] == ‘0‘)
25     {
26         ret.erase(0, 1);
27     }
28     return make_pair(ret, rem); //返回<商，余数>
29 
30 }

 1 string mul(string num1, string num2)
 2 {
 3     //简化乘法的过程
 4     if (num2.size() > num1.size())
 5     {
 6         swap(num2, num1);
 7     }
 8     string ret = "0";
 9     for (int i = num2.size() - 1; i >= 0; i++)
10     {
11         //获取当前位乘数的值
12         int curDigit = num2[i] - ‘0‘;
13         int step = 0;
14         //当前位乘积的结果
15         string tmp = num1;
16         for (int j = tmp.size(); j >= 0; j--)
17         {
18             //逐位相乘
19             tmp[j] = (tmp[j] - ‘0‘) * curDigit + step;
20             //更新进位
21             if (tmp[j] > 9)
22             {
23                 step = tmp[j] / 10;
24                 tmp = tmp[j] % 10;
25             }
26             else step = 0;
27             tmp[j] += ‘0‘;//还原成字符
28         }
29         //最后判断是否需要进位
30         if (step > 0)
31         {
32             tmp.insert(0, 1, step + ‘0‘);
33         }
34         //补零
35         tmp.append(num2.size() - 1 - i, ‘0‘);
36         //累加一次乘法的结果
37         ret = add(ret, tmp);
38     }
39     return ret;
40 }