C++ 高精度算法及N的阶乘

所谓高精度就是用普通类型计算都会溢出的大数运算
高精度算法在做题时经常遇到且经常性的模板化，这里做一下总结

以下的程序重载了高精度中可能遇到的多种运算符，但不能出现负数

#include
#include
#include
using namespace std;
//输出数据最大长度，根据情况更改大小，不要太大
const int maxn = 50000;
struct bign
{
  int len, s[maxn];

  bign()    //初始化构造函数
  {
    memset(s, 0, sizeof(s));
    len = 1;
  }
 //构造函数
  bign(int num)
  {
    *this = num;
  }
  //构造函数
  bign(const char* num)
  {
    *this = num;
  }
  //重载int=号
  bign operator = (int num)
  {
    char s[maxn];
    sprintf(s, "%d", num);
    *this = s;
    return *this;
  }
  //重载字符型=
  bign operator = (const char* num)
  {
    len = strlen(num);
    for(int i = 0; i < len; i++) s[i] = num[len-i-1] - '0';
    return *this;
  }
  //将数组s转化为字符串
  string str() const
  {
    string res = "";
    for(int i = 0; i < len; i++) res = (char)(s[i] + '0') + res;
    if(res == "") res = "0";
    return res;
  }
  //重载+
  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;
  }
  //去除数据的前导0
  void clean()
  {
    while(len > 1 && !s[len-1]) len--;
  }
  //重载*
  bign operator * (const bign& b)
  {
    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;
  }
  //重载-,不支持负数
  bign operator - (const bign& b)
  {
    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;
  }
  //重载比较运算符
  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)
  {
    return !(b > *this);
  }

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

  bign operator += (const bign& b)
  {
    *this = *this + b;
    return *this;
  }
};
  //重载输入流
istream& operator >> (istream &in, bign& x)
{
  string s;
  in >> s;
  x = s.c_str();
  return in;
}
  //重载输出流
ostream& operator << (ostream &out, const bign& x)
{
  out << x.str();
  return out;
}

int main()
{
    bign a;
    while(cin>>a>>b)
    {
        cout<

下面是计算N的阶乘的程序的简单写法，可以根据个人情况加入上面程序的构造函数

注意：2e5的阶乘将会超过50000位，maxn根据自己情况修改大小

#include
#include
#include
using namespace std;
const int maxn=50000;
struct node
{
    int f[maxn],len;
    node()
    {
        memset(f,0,sizeof(f));
        len=1;
    }
    node operator=(const string&p)
    {
        len=p.size();
        for(int i=0;i>(istream&in,node&p)
{
    string s;
    in>>s;
    p=s;
    return in;
}
ostream&operator<<(ostream&out,const node&p)
{
    for(int i=p.len-1;i>=0;i--)
        out<>n)
    {
        node a,t;
        a=1;
        for(int i=1;i<=n;i++)
        {
            t=i;
            a=a*t;
        }
        cout<