2的幂位数大整数分治算法

标签：2的幂位数大整数分治算法

#include <iostream>
#include <cstring>
#include <string>
#include <cstdio>

using namespace std;
//500 digits at most
struct Num{
    int num[1000],len;
    Num(){
        memset(num,0,sizeof(num));
        len=1;
    }
    Num(const string &s){
        len=s.size();
        for(int i=0;i<len;++i){
            num[i]=s[len-1-i]-'0';
        }
    }
    Num& operator=(const Num& right){
        copy(right.num,right.num+1000,this->num);
        len=right.len;
        return *this;
    }
    friend ostream& operator<<(ostream &os,const Num &output){
        for(int i=output.len-1;i>=0;--i)
            os<<output.num[i];
        return os;
    }
    friend Num operator+(Num &x,Num &y){
        Num ans;
        ans.len=max(x.len,y.len);
        for(int i=0;i<ans.len;++i){
            ans.num[i]=x.num[i]+y.num[i];
        }
        for(int i=0;i<ans.len;++i){
            ans.num[i+1]+=ans.num[i]/10;
            ans.num[i]%=10;
        }
        if(ans.num[ans.len]) ++ans.len;
        return ans;
    }
    friend Num operator-(const Num &left,const Num& right){
        Num ans;
        ans.len=max(left.len,right.len);
        for(int i=0;i<ans.len;++i)
            ans.num[i]=left.num[i]-right.num[i];
        for(int i=0;i<ans.len;++i){
            if(ans.num[i]<0){
                --ans.num[i+1];
                ans.num[i]+=10;
            }
        }
        while(ans.len>=1 && !ans.num[ans.len-1]) --ans.len;
        return ans;
    }
    friend Num mul(const Num &x,const Num &y){
        if(x.len==1 && y.len==1) return to_string(x.num[0]*y.num[0]);
        Num a,b,c,d,ans;
        int maxlen=max(x.len,y.len),len1=maxlen>>1;
        copy(x.num,x.num+len1,b.num);
        b.len=len1;
        copy(x.num+len1,x.num+maxlen,a.num);
        a.len=maxlen-len1;
        copy(y.num,y.num+len1,d.num);
        d.len=len1;
        copy(y.num+len1,y.num+maxlen,c.num);
        c.len=maxlen-len1;
        Num tem_ac=mul(a,c),tem_bd=mul(b,d),ac,bd,fin_cdab;
        Num cdab=mul(c+d,a+b)-tem_ac-tem_bd;
        copy(tem_ac.num,tem_ac.num+tem_ac.len,ac.num+len1*2);
        ac.len=tem_ac.len+2*len1;
        copy(cdab.num,cdab.num+cdab.len,fin_cdab.num+len1);
        fin_cdab.len=cdab.len+len1;
        ans=ac+fin_cdab;
        ans=ans+tem_bd;
        while(ans.len>1 && !ans.num[ans.len-1]) --ans.len;
        return ans;
    }
    friend Num operator*(const Num &x,const Num &y){
        return mul(x,y);
    }
};

int main()
{
    string a,b;
    while(cin>>a>>b){
        Num x=a,y=b;
        cout<<"***************************************************"<<endl;
        cout<<a<<endl<<"*"<<endl<<b<<endl<<"="<<endl<<x*y<<endl;
        cout<<"***************************************************"<<endl;
    }
    return 0;
}

2的幂位数大整数分治算法

标签：2的幂位数大整数分治算法

原文地址：http://blog.csdn.net/fangpinlei/article/details/45030423