Apple HDU6206

2018TYUT暑期ACM模拟赛(8)
​​​Apple HDU6206 ​​​
题意：给出四个点，问第四个点在不在前三个点构成的圆内。
思路：一开始队里的说这个简单，妈耶，后来大家发现这个小技巧是不行的，还是要高精度，作为模拟赛的所以默默去找了一个高精度大数模板。有点小不理解自己套上圆心半径计算公式的模板不对，然后看了别的大佬的题解，为啥是求半径的公式是两点相加的平法？？？？？有点小蒙。

#include<string>
#include<iostream>
#include<iosfwd>
#include<cmath>
#include<cstring>
#include<stdlib.h>
#include<stdio.h>
#include<cstring>
#define MAX_L 2005 //最大长度，可以修改
using namespace std;
#define max_n 100000000//q去除小树
class bign
{
public:
    int len, s[MAX_L];//数的长度，记录数组
//构造函数
    bign();
    bign(const char*);
    bign(int);
    bool sign;//符号 1正数 0负数
    string toStr() const;//转化为字符串，主要是便于输出
    friend istream& operator>>(istream &,bign &);//重载输入流
    friend ostream& operator<<(ostream &,bign &);//重载输出流
//重载复制
    bign operator=(const char*);
    bign operator=(int);
    bign operator=(const string);
//重载各种比较
    bool operator>(const bign &) const;
    bool operator>=(const bign &) const;
    bool operator<(const bign &) const;
    bool operator<=(const bign &) const;
    bool operator==(const bign &) const;
    bool operator!=(const bign &) const;
//重载四则运算
    bign operator+(const bign &) const;
    bign operator++();
    bign operator++(int);
    bign operator+=(const bign&);
    bign operator-(const bign &) const;
    bign operator--();
    bign operator--(int);
    bign operator-=(const bign&);
    bign operator*(const bign &)const;
    bign operator*(const int num)const;
    bign operator*=(const bign&);
    bign operator/(const bign&)const;
    bign operator/=(const bign&);
//四则运算的衍生运算
    bign operator%(const bign&)const;//取模（余数）
    bign factorial()const;//阶乘
    bign Sqrt()const;//整数开根（向下取整）
    bign pow(const bign&)const;//次方
//一些乱乱的函
    void clean();
    ~bign();
};
#define max(a,b) a>b ? a : b
#define min(a,b) a<b ? a : b

bign::bign()
{
    memset(s, 0, sizeof(s));
    len = 1;
    sign = 1;
}
bign::bign(const char *num)
{
    *this = num;
}
bign::bign(int num)
{
    *this = num;
}
string bign::toStr() const
{
    string res;
    res = "";
    for (int i = 0; i < len; i++)
        res = (char)(s[i] + '0') + res;
    if (res == "")
        res = "0";
    if (!sign&&res != "0")
        res = "-" + res;
    return res;
}
istream &operator>>(istream &in, bign &num)
{
    string str;
    in>>str;
    num=str;
   return in;
}
ostream &operator<<(ostream &out, bign &num)
{
    out<<num.toStr();
    return out;
}
bign bign::operator=(const char *num)
{
    memset(s, 0, sizeof(s));
    char a[MAX_L] = "";
    if (num[0] != '-')
        strcpy(a, num);
    else
        for (int i = 1; i < strlen(num); i++)
            a[i - 1] = num[i];
    sign = !(num[0] == '-');
    len = strlen(a);
    for (int i = 0; i < strlen(a); i++)
        s[i] = a[len - i - 1] - 48;
    return *this;
}

bign bign::operator=(int num)
{
   if (num < 0)
        sign = 0, num = -num;
   else
        sign = 1;
    char temp[MAX_L];
    sprintf(temp, "%d", num);
    *this = temp;
    return *this;
}
bign bign::operator=(const string num)
{
    const char *tmp;
    tmp = num.c_str();
    *this = tmp;
    return *this;
}
bool bign::operator<(const bign &num) const
{
    if (sign^num.sign)
        return num.sign;
    if (len != num.len)
        return len < num.len;
    for (int i = len - 1; i >= 0; i--)
        if (s[i] != num.s[i])
            return sign ? (s[i] < num.s[i]) : (!(s[i] < num.s[i]));
    return !sign;
}
bool bign::operator>(const bign&num)const
{
    return num < *this;
}
bool bign::operator<=(const bign&num)const
{
     return !(*this>num);
}
bool bign::operator>=(const bign&num)const
{
    return !(*this<num);
}

bool bign::operator!=(const bign&num)const
{
    return *this > num || *this < num;
}

bool bign::operator==(const bign&num)const
{
    return !(num != *this);
}
bign bign::operator+(const bign &num) const
{
    if (sign^num.sign)
    {
        bign tmp = sign ? num : *this;
        tmp.sign = 1;
        return sign ? *this - tmp : num - tmp;
    }
    bign result;
    result.len = 0;
    int temp = 0;
    for (int i = 0; temp || i < (max(len, num.len)); i++)
    {
        int t = s[i] + num.s[i] + temp;
        result.s[result.len++] = t % 10;
        temp = t / 10;
    }
    result.sign = sign;
    return result;
}
bign bign::operator++()
{
    *this = *this + 1;
    return *this;
}
bign bign::operator++(int)
{
    bign old = *this;
    ++(*this);
    return old;
}
bign bign::operator+=(const bign &num)
{
    *this = *this + num;
    return *this;
}
bign bign::operator-(const bign &num) const
{
    bign b=num,a=*this;
    if (!num.sign && !sign)
    {
        b.sign=1;
        a.sign=1;
        return b-a;
    }
    if (!b.sign)
    {
        b.sign=1;
        return a+b;
    }
    if (!a.sign)
    {
        a.sign=1;
        b=bign(0)-(a+b);
        return b;
    }
    if (a<b)
    {
        bign c=(b-a);
        c.sign=false;
        return c;
    }
    bign result;
    result.len = 0;
    for (int i = 0, g = 0; i < a.len; i++)
    {
        int x = a.s[i] - g;
        if (i < b.len) x -= b.s[i];
        if (x >= 0) g = 0;
        else
        {
            g = 1;
            x += 10;
        }
        result.s[result.len++] = x;
    }
    result.clean();
    return result;
}
bign bign::operator * (const bign &num)const
{
    bign result;
    result.len = len + num.len;
    for (int i = 0; i < len; i++)
        for (int j = 0; j < num.len; j++)
            result.s[i + j] += s[i] * num.s[j];
    for (int i = 0; i < result.len; i++)
    {
        result.s[i + 1] += result.s[i] / 10;
        result.s[i] %= 10;
    }
    result.clean();
    result.sign = !(sign^num.sign);
    return result;
}
bign bign::operator*(const int num)const
{
    bign x = num;
    bign z = *this;
    return x*z;
}
bign bign::operator*=(const bign&num)
{
    *this = *this * num;
    return *this;
}

bign bign::operator /(const bign&num)const
{
   bign ans;
    ans.len = len - num.len + 1;
    if (ans.len < 0)
    {
        ans.len = 1;
        return ans;
    }
    bign divisor = *this, divid = num;
    divisor.sign = divid.sign = 1;
    int k = ans.len - 1;
    int j = len - 1;
    while (k >= 0)
    {
        while (divisor.s[j] == 0) j--;
        if (k > j) k = j;
        char z[MAX_L];
        memset(z, 0, sizeof(z));
        for (int i = j; i >= k; i--)
            z[j - i] = divisor.s[i] + '0';
        bign dividend = z;
        if (dividend < divid) { k--; continue; }
        int key = 0;
        while (divid*key <= dividend) key++;
        key--;
        ans.s[k] = key;
        bign temp = divid*key;
        for (int i = 0; i < k; i++)
            temp = temp * 10;
        divisor = divisor - temp;
        k--;
    }
    ans.clean();
    ans.sign = !(sign^num.sign);
    return ans;
}

bign bign::operator/=(const bign&num)
{
    *this = *this / num;
    return *this;
}

bign bign::operator%(const bign& num)const
{
    bign a = *this, b = num;
    a.sign = b.sign = 1;
    bign result, temp = a / b*b;
    result = a - temp;
    result.sign = sign;
    return result;

}

bign bign::pow(const bign& num)const
{

    bign result = 1;

    for (bign i = 0; i < num; i++)
        result = result*(*this);
    return result;
}
bign bign::factorial()const
{
    bign result = 1;
    for (bign i = 1; i <= *this; i++)
        result *= i;
    return result;
}
void bign::clean()
{
    if (len == 0) len++;
    while (len > 1 && s[len - 1] == '\0')
        len--;
}

bign bign::Sqrt()const
{
    if(*this<0)return -1;
    if(*this<=1)return *this;
    bign l=0,r=*this,mid;
    while(r-l>1)
    {
        mid=(l+r)/2;
        if(mid*mid>*this)
            r=mid;
        else 
            l=mid;
    }
    return l;
}
bign::~bign()
{
}

bign X,Y;
 struct Point{
    bign x,y;
};
bign dis(Point x,Point y)
{
    bign tmp=(x.x+y.x)*(x.x+y.x)+(x.y+y.y)*(x.y+y.y);
    return tmp;//不sqrt
}
//爆了
void solve(Point a,Point b,Point c)//三点共圆圆心公式
{
    bign e,f,g,h,i,j;

    e=(a.x*a.x-b.x*b.x+a.y*a.y-b.y*b.y)/2;
    f=(a.x*a.x-c.x*c.x+a.y*a.y-c.y*c.y)/2;

    g=a.x-b.x;
    h=a.y-b.y;
    i=a.x-c.x;
    j=a.y-c.y;    

    X=(e*j-h*f)/(h*i-g*j);
    Y=(g*f-i*e)/(h*i-g*j);

   // bign tmp=(X+a.x)*(X+a.x)+(Y+a.y)*(Y+a.y);
    //R=tmp; //不sqrt了
}
int main()
{
    int t;
    cin>>t;
    while(t--)
    {
        Point x,y,z,tmp;
        cin>>x.x>>x.y>>y.x>>y.y>>z.x>>z.y>>tmp.x>>tmp.y;
        x.x*=max_n;x.y*=max_n;
        y.x*=max_n;y.y*=max_n;
        z.x*=max_n;z.y*=max_n;
        tmp.x*=max_n;tmp.y*=max_n;
        solve(x,y,z);//
        //x.x=X;y.y=Y;
        bign R=(X+x.x)*(X+x.x)+(Y+x.y)*(Y+x.y);
        bign dist=(X+tmp.x)*(X+tmp.x)+(Y+tmp.y)*(Y+tmp.y);
        if(R<dist) printf("Accepted\n");//圈外
        else printf("Rejected\n");//圈nei
       } 
    return 0;
}