[SDOI2009]SuperGCD

标签：while   lin   ike   name   tor   math   大小   make   include   

【题面】：
[SDOI2009]SuperGCD

【思路】：
毒瘤高精。。
考这种题真不知道出题人怎么想的，高精就算了还要压八位。。我高精板子都挂了还是寻欢大神给我了个板子\(qwq\)

这是一道裸（du）的（liu）\(GCD\)，当你把一切运算符都重载之后，你就可以愉快地\(coding\)出来\(gcd\)！then TLE

你还需要这个：更相减损术

可半者半之，不可半者，副置分母、子之数，以少减多，更相减损，求其等也。以等数约之。\(orz\)

具体来说，就是（如果需要对分数进行约分，那么）可以折半的话，就折半（也就是用\(2\)来约分）。如果不可以折半的话，那么就比较分母和分子的大小，用大数减去小数，互相减来减去，一直到减数与差相等为止，用这个相等的数字来约分。

或者用\(Stein\)算法也可，都是神仙\(orz\)

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;

struct huge{
    #define N_huge 10000
    #define base 100000000
    static char s[N_huge*10];
    typedef long long value;
    value a[N_huge];int len;
    void clear(){len=1;a[len]=0;}
    huge(){clear();}
    huge(value x){*this=x;}
    huge(char s[]){this->str(s);}
    huge operator =(const huge &b){
        len=b.len;for (int i=1;i<=len;++i)a[i]=b.a[i]; return *this;
    }
    huge operator +(const huge &b){
        int L=len>b.len?len:b.len;huge tmp;
        for (int i=1;i<=L+1;++i)tmp.a[i]=0;
        for (int i=1;i<=L;++i){
            if (i>len)tmp.a[i]+=b.a[i];
            else if (i>b.len)tmp.a[i]+=a[i];
            else {
                tmp.a[i]+=a[i]+b.a[i];
                if (tmp.a[i]>=base){
                    tmp.a[i]-=base;++tmp.a[i+1];
                }
            }
        }
        if (tmp.a[L+1])tmp.len=L+1;
            else tmp.len=L;
        return tmp;
    }
    huge operator -(huge b){
        int L=len>b.len?len:b.len;huge tmp;
        for (int i=1;i<=L+1;++i)tmp.a[i]=0;
        for (int i=1;i<=L;++i){
            if (i>b.len)b.a[i]=0;
            tmp.a[i]+=a[i]-b.a[i];
            if (tmp.a[i]<0){
                tmp.a[i]+=base;--tmp.a[i+1];
            }
        }
        while (L>1&&!tmp.a[L])--L;
        tmp.len=L;
        return tmp;
    }
    huge operator *(const huge &b)const{
        int L=len+b.len;huge tmp;
        for (int i=1;i<=L;++i)tmp.a[i]=0;
        for (int i=1;i<=len;++i)
            for (int j=1;j<=b.len;++j){
                tmp.a[i+j-1]+=a[i]*b.a[j];
                if (tmp.a[i+j-1]>=base){
                    tmp.a[i+j]+=tmp.a[i+j-1]/base;
                    tmp.a[i+j-1]%=base;
                }
            }
        tmp.len=len+b.len;
        while (tmp.len>1&&!tmp.a[tmp.len])--tmp.len;
        return tmp;
    }
    pair<huge,huge> divide(const huge &a,const huge &b){
        int L=a.len;huge c,d;
        for (int i=L;i;--i){
            c.a[i]=0;d=d*base;d.a[1]=a.a[i];
            //while (d>=b){d-=b;++c.a[i];}
            int l=0,r=base-1,mid;
            while (l<r){
                mid=(l+r+1)>>1;
                if (b*mid<=d)l=mid;
                    else r=mid-1;
            }
            c.a[i]=l;d-=b*l;
        }
        while (L>1&&!c.a[L])--L;c.len=L;
        return make_pair(c,d);
    }
    huge operator /(value x){
        value d=0;huge tmp;
        for (int i=len;i;--i){
            d=d*base+a[i];
            tmp.a[i]=d/x;d%=x;
        }
        tmp.len=len;
        while (tmp.len>1&&!tmp.a[tmp.len])--tmp.len;
        return tmp;
    }
    value operator %(value x){
        value d=0;
        for (int i=len;i;--i)d=(d*base+a[i])%x;
        return d;
    }
    huge operator /(const huge &b){return divide(*this,b).first;}
    huge operator %(const huge &b){return divide(*this,b).second;}
    huge &operator +=(const huge &b){*this=*this+b;return *this;}
    huge &operator -=(const huge &b){*this=*this-b;return *this;}
    huge &operator *=(const huge &b){*this=*this*b;return *this;}
    huge operator /=(const huge &b){*this=*this/b;return *this;}
    huge operator %=(const huge &b){*this=*this%b;return *this;}
    huge &operator ++(){huge T;T=1;*this=*this+T;return *this;}
    huge &operator --(){huge T;T=1;*this=*this-T;return *this;}
    huge operator ++(int){huge T,tmp=*this;T=1;*this=*this+T;return tmp;}
    huge operator --(int){huge T,tmp=*this;T=1;*this=*this-T;return tmp;}
    huge operator +(value x){huge T;T=x;return *this+T;}
    huge operator -(value x){huge T;T=x;return *this-T;}
    huge operator *(value x){huge T;T=x;return *this*T;}
    //huge operator /(value x){huge T;T=x;return *this/T;}
    //huge operator %(value x){huge T;T=x;return *this%T;}
    huge operator *=(value x){*this=*this*x;return *this;}
    huge operator +=(value x){*this=*this+x;return *this;}
    huge operator -=(value x){*this=*this-x;return *this;}
    huge operator /=(value x){*this=*this/x;return *this;}
    huge operator %=(value x){*this=*this%x;return *this;}
    bool operator ==(value x){huge T;T=x;return *this==T;}
    bool operator !=(value x){huge T;T=x;return *this!=T;}
    bool operator <=(value x){huge T;T=x;return *this<=T;}
    bool operator >=(value x){huge T;T=x;return *this>=T;}
    bool operator <(value x){huge T;T=x;return *this<T;}
    bool operator >(value x){huge T;T=x;return *this>T;}
    huge operator =(value x){
        len=0;
        while (x)a[++len]=x%base,x/=base;
        if (!len)a[++len]=0;
        return *this;
    }
    bool operator <(const huge &b){
        if (len<b.len)return 1;
        if (len>b.len)return 0;
        for (int i=len;i;--i){
            if (a[i]<b.a[i])return 1;
            if (a[i]>b.a[i])return 0;
        }
        return 0;
    }
    bool operator ==(const huge &b){
        if (len!=b.len)return 0;
        for (int i=len;i;--i)
            if (a[i]!=b.a[i])return 0;
        return 1;
    }
    bool operator !=(const huge &b){return !(*this==b);}
    bool operator >(const huge &b){return !(*this<b||*this==b);}
    bool operator <=(const huge &b){return (*this<b)||(*this==b);}
    bool operator >=(const huge &b){return (*this>b)||(*this==b);}
    void str(char s[]){
        int l=strlen(s);value x=0,y=1;len=0;
        for (int i=l-1;i>=0;--i){
            x=x+(s[i]-'0')*y;y*=10;
            if (y==base)a[++len]=x,x=0,y=1;
        }
        if (!len||x)a[++len]=x;
    }
    void read(){
        scanf("%s",s);this->str(s);
    }
    void print(){
        printf("%d",(int)a[len]);
        for (int i=len-1;i;--i){
            for (int j=base/10;j>=10;j/=10){
                if (a[i]<j)printf("0");
                    else break;
            }
            printf("%d",(int)a[i]);
        }
        printf("\n");
    }
};char huge::s[N_huge*10];

huge gcd(huge a,huge b){
    
    huge c;c.clear();
    c = 1;
    while(((a%2)==0)&&((b%2)==0)){
        a/=2;
        b/=2;
        c*=2;
    }
    while(a!=b){
        if(a>b)a-=b;
        else b-=a;
    }
    c*=a;
    return c;
}

huge a;
huge b;
huge ans;

int main(){
    a.read();b.read();
    ans = gcd(a,b);
    ans.print();
}

[SDOI2009]SuperGCD

标签：while   lin   ike   name   tor   math   大小   make   include   

原文地址：https://www.cnblogs.com/lajioj/p/9562696.html