UVA 10106 Product 简单高精度乘法

题目要求输入两个非负的整数后求出他们的积

两个数的范围可以超出 long long型

第一次做高精度的题，这是我自己写的AC代码：

用字符串模拟乘法运算：

Result  :  Accepted     Memory  :  0 KB     Time :  16 ms

/* * Author: Gatevin * Created Time:  2014/7/3 16:12:18 * File Name: test.cpp */#include<iostream>#include<sstream>#include<fstream>#include<vector>#include<list>#include<deque>#include<queue>#include<stack>#include<map>#include<set>#include<bitset>#include<algorithm>#include<cstdio>#include<cstdlib>#include<cstring>#include<cctype>#include<cmath>#include<ctime>#include<iomanip>using namespace std;const double eps(1e-8);typedef long long lint;string s1,s2;int main(){   while(cin>>s1>>s2)   {       if(s1.length() == 1 && s1[0] == '0')//处理输入有零的情况       {           cout<<"0"<<endl;           continue;       }       if(s2.length() == 1 && s2[0] == '0')       {           cout<<"0"<<endl;           continue;       }       reverse(s1.begin(), s1.end());//先将数字从末尾到首位摆放       reverse(s2.begin(), s2.end());       string ans(s1.length() + s2.length() + 1, '0');//两个整数相乘，积的位数不会超过他们的位数和       int c = 0;       for(int i = 0; i < s1.length(); i++)//模拟乘法运算       {           c = 0;           for(int j = 0; j < s2.length(); j++)           {              /*               *一个数的倒数第i位乘上另一个数的倒数第j位改变的是倒数第i + j位(i,j是编号，从0开始)               */               int tmp1 =  ((ans[i + j] - '0') + (s1[i] - '0')*(s2[j] - '0') + c) / 10;               int tmp2 =  ((ans[i + j] - '0') + (s1[i] - '0')*(s2[j] - '0') + c) % 10;               ans[i + j] = tmp2 + '0';               c = tmp1;           }           ans[i + s2.length()] += c;       }       int flag;       for(int i = ans.length() - 1; i >= 0; i--)//输出忽略前导零       {           if(ans[i] != '0')           {               flag = i;               break;           }       }       for(int i = flag; i >= 0; i--)       {           cout<<ans[i];       }       cout<<endl;   }   return 0;}

这是之后使用高精度运算模板的代码：

Result  :  Accepted   Memory  : 0 KB   Time  :  13 ms

/* * Author: Gatevin * Created Time:  2014/7/4 12:58:26 * File Name: test.cpp */#include<iostream>#include<sstream>#include<fstream>#include<vector>#include<list>#include<deque>#include<queue>#include<stack>#include<map>#include<set>#include<bitset>#include<algorithm>#include<cstdio>#include<cstdlib>#include<cstring>#include<cctype>#include<cmath>#include<ctime>#include<iomanip>using namespace std;const double eps(1e-8);typedef long long lint;struct bign{    int len,s[1000];    bign()    {        len = 0;        memset(s, 0 ,sizeof(s));    }        bign operator = (const char* num)    {        len = strlen(num);        for(int i = 0; i < len; i++)        {            s[i] = num[len - i - 1] - '0';        }        return *this;    }        bign operator = (int num)    {        char s[1000];        sprintf(s, "%d", num);        *this = s;        return *this;    }        bign(int num)    {        *this = num;    }        bign(const char* num)    {        *this = num;    }        string str() const    {        string res = "";        for(int i = 0; i < len; i++)        {            res = (char)(s[i] + '0') + res;        }        if(res == "")        {            res = "0";        }        return res;    }        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;    }        bign operator * (const bign& b) const    {        bign c;        c.len = len + b.len;        int k;        for(int i = 0; i < len; i++)        {            k = 0;            for(int j = 0; j < b.len; j++)            {                int tmp1 = (s[i] * b.s[j] + k + c.s[i + j]) / 10;                int tmp2 = (s[i] * b.s[j] + k + c.s[i + j]) % 10;                c.s[i + j] = tmp2;                k = tmp1;            }            c.s[i + b.len] += k;        }        while(c.len > 1)        {            if(c.s[c.len - 1] == 0)            {                c.len--;            }            else            {                break;            }        }        return c;    }    bign operator += (const bign& b)    {        *this = *this + b;        return *this;    }        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;    }};    istream& operator >> (istream &in, bign & x)    {        string s;        in >> s;        x = s.c_str();        return in;    }        ostream& operator << (ostream & out, const bign& x)    {        out << x.str();        return out;    }int main(){    bign x1,x2;   while(cin>>x1>>x2)  cout<<x1*x2<<endl;     return 0;}

模板代码我没加注释，不懂的话可以看看我之后整理的 bign 模板