模板 — 高精度计算

Posted 2023-01-11 繁凡さん

ACM-ICPC在线模板

一套重载运算符的高精模板

#include<iostream>
#include<string>
#include<cstring>
#include<cstdio>
using namespace std;
const int N = 1005;
struct bign

    int len,s[N];
    bign()    memset(s,0,sizeof(s));  len=1;  
    bign(int num)    *this=num; 
    bign(char *num)  *this=num; 
    bign operator =(int num)
    
        char c[N];
        sprintf(c,"%d",num);
        *this=c;
        return *this;
    
    bign operator =(const char *num)
    
        len=strlen(num);
        for (int i=0;i<len;i++) s[i]=num[len-1-i]-'0';
        return *this;
    
    string str()
    
        string res="";
        for (int i=0;i<len;i++) res=(char)(s[i]+'0')+res;
        return res;
    
    void clean()
    
        while (len>1&&!s[len-1]) len--;
    
    bign operator +(const bign &b)
    
        bign c;    
        c.len=0;
        for (int i=0,g=0;g||i<len||i<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)
    
        bign c;
        c.len=0;
        int x;     
        for (int i=0,g=0;i<len;i++)
        
            x=s[i]-g;
            if (i<b.len) x-=b.s[i];
            if (x>=0) g=0;
            else          
                x+=10;
                g=1;
            ;
            c.s[c.len++]=x;
        
        c.clean();
        return c;
    
    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;  
    
    bool operator <(const bign &b)
    
        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;
    
    bign operator +=(const bign &b)
    
        *this=*this+b;
        return *this;
    
    bign operator -=(const bign &b)
    
        *this=*this-b;
        return *this;
      
;
istream& operator >>(istream &in,bign &x)

  string s;
  in>>s;
  x=s.c_str();
  return in;

ostream& operator <<(ostream &out,bign &x)

    out<<x.str();
    return out;

int main()
    bign a,b,c;
    ios::sync_with_stdio(false);
    cin>>a>>b;
//    cout<<a<<endl;
//    cout<<b<<endl;
    c=a+b;
    cout<<c<<endl;
    return 0;