UVa 10106 Product：高精度

10106 - Product

Time limit: 3.000 seconds

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=97&page=show_problem&problem=1047

The Problem

The problem is to multiply two integers X, Y. (0<=X,Y<10250)
The Input

The input will consist of a set of pairs of lines. Each line in pair contains one multiplyer.
The Output

For each input pair of lines the output line should consist one integer the product.
Sample Input

12
12
2
222222222222222222222222

Sample Output

144
444444444444444444444444

完整代码：

/*0.012s*/

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn = 505;  

char numstr[maxn];  

struct bign
{
    int len, s[maxn];  

    bign()
    {
        memset(s, 0, sizeof(s));
        len = 1;
    }  

    bign(int num)
    {
        *this = num;
    }  

    bign(const char* num)
    {
        *this = num;
    }  

    bign operator = (const int num)
    {
        char s[maxn];
        sprintf(s, "%d", num);
        *this = s;
        return *this;
    }  

    bign operator = (const char* num)
    {
        len = strlen(num);
        for (int i = 0; i < len; i++) s[i] = num[len - i - 1] & 15;
        return *this;
    }  

    ///输出
    const char* str() const
    {
        if (len)
        {
            for (int i = 0; i < len; i++)
                numstr[i] = '0' + s[len - i - 1];
            numstr[len] = '\0';
        }
        else strcpy(numstr, "0");
        return numstr;
    }  

    ///去前导零
    void clean()
    {
        while (len > 1 && !s[len - 1]) len--;
    }  

    ///加
    bign operator + (const bign& b) const
    {
        bign c;
        c.len = 0;
        for (int i = 0, g = 0; g || i < max(len, b.len); i++)
        {
            int x = g;
            if (i < len) x += s[i];
            if (i < b.len) x += b.s[i];
            c.s[c.len++] = x % 10;
            g = x / 10;
        }
        return c;
    }  

    ///减
    bign operator - (const bign& b) const
    {
        bign c;
        c.len = 0;
        for (int i = 0, g = 0; i < len; i++)
        {
            int x = s[i] - g;
            if (i < b.len) x -= b.s[i];
            if (x >= 0) g = 0;
            else
            {
                g = 1;
                x += 10;
            }
            c.s[c.len++] = x;
        }
        c.clean();
        return c;
    }  

    ///乘
    bign operator * (const bign& b) const
    {
        bign c;
        c.len = len + b.len;
        for (int i = 0; i < len; i++)
            for (int j = 0; j < b.len; j++)
                c.s[i + j] += s[i] * b.s[j];
        for (int i = 0; i < c.len - 1; i++)
        {
            c.s[i + 1] += c.s[i] / 10;
            c.s[i] %= 10;
        }
        c.clean();
        return c;
    }  

    ///除
    bign operator / (const bign &b) const
    {
        bign ret, cur = 0;
        ret.len = len;
        for (int i = len - 1; i >= 0; i--)
        {
            cur = cur * 10;
            cur.s[0] = s[i];
            while (cur >= b)
            {
                cur -= b;
                ret.s[i]++;
            }
        }
        ret.clean();
        return ret;
    }  

    ///模、余
    bign operator % (const bign &b) const
    {
        bign c = *this / b;
        return *this - c * b;
    }  

    bool operator < (const bign& b) const
    {
        if (len != b.len) return len < b.len;
        for (int i = len - 1; i >= 0; i--)
            if (s[i] != b.s[i]) return s[i] < b.s[i];
        return false;
    }  

    bool operator > (const bign& b) const
    {
        return b < *this;
    }  

    bool operator <= (const bign& b) const
    {
        return !(b < *this);
    }  

    bool operator >= (const bign &b) const
    {
        return !(*this < b);
    }  

    bool operator == (const bign& b) const
    {
        return !(b < *this) && !(*this < b);
    }  

    bool operator != (const bign &a) const
    {
        return *this > a || *this < a;
    }  

    bign operator += (const bign &a)
    {
        *this = *this + a;
        return *this;
    }  

    bign operator -= (const bign &a)
    {
        *this = *this - a;
        return *this;
    }  

    bign operator *= (const bign &a)
    {
        *this = *this * a;
        return *this;
    }  

    bign operator /= (const bign &a)
    {
        *this = *this / a;
        return *this;
    }  

    bign operator %= (const bign &a)
    {
        *this = *this % a;
        return *this;
    }
} a, b;  

int main(void)
{
    char ch;
    while (~(ch = getchar()))
    {
        ungetc(ch, stdin);
        a = gets(numstr), b = gets(numstr);
        puts((a * b).str());
    }
    return 0;
}

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