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<cstdio>#include<cstring>#include<iostream>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<<a.len<<endl;        cout<<a<<endl;        cout<<a+a<<endl;        cout<<a*a<<endl;    }    return 0;}

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<iostream>#include<cstdio>#include<cstring>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<len;i++)            f[len-i-1]=p[i]-'0';        return *this;    }    node operator=(int n)    {        if(n==0)return*this;        int i=0;        while(n)        {            f[i++]=n%10;            n/=10;        }        len=i;        return*this;    }    node operator*(const node&p)    {        node tmp;        int y=0;        tmp.len=len+p.len;        for(int i=0;i<len;i++)            for(int j=0;j<p.len;j++)            tmp.f[i+j]+=f[i]*p.f[j];        for(int i=0;i<tmp.len;i++)        {            tmp.f[i+1]+=tmp.f[i]/10;            tmp.f[i]%=10;        }        if(!tmp.f[tmp.len-1])tmp.len--;        return tmp;    }};istream&operator>>(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<<p.f[i];    return out;}int main(){    //freopen("in.txt","r",stdin);    //freopen("out.txt","w",stdout);    int n;    while(cin>>n)    {        node a,t;        a=1;        for(int i=1;i<=n;i++)        {            t=i;            a=a*t;        }        cout<<a.len<<endl;        cout<<a<<endl;    }    return 0;}