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  • 2017 ACM/ICPC Asia Regional Qingdao Online 青岛区域赛

    1001Apple

    Problem Description
    Apple is Taotao's favourite fruit. In his backyard, there are three apple trees with coordinates (x1,y1), (x2,y2), and (x3,y3). Now Taotao is planning to plant a new one, but he is not willing to take these trees too close. He believes that the new apple tree should be outside the circle which the three apple trees that already exist is on. Taotao picked a potential position (x,y) of the new tree. Could you tell him if it is outside the circle or not?
     
    
    Input
    The first line contains an integer T, indicating that there are T(T≤30) cases.
    In the first line of each case, there are eight integers x1,y1,x2,y2,x3,y3,x,y, as described above.
    The absolute values of integers in input are less than or equal to 1,000,000,000,000.
    It is guaranteed that, any three of the four positions do not lie on a straight line.
     
    
    Output
    For each case, output "Accepted" if the position is outside the circle, or "Rejected" if the position is on or inside the circle.
     
    
    Sample Input
    3
    -2 0 0 -2 2 0 2 -2
    -2 0 0 -2 2 0 0 2
    -2 0 0 -2 2 0 1 1
     
    
    Sample Output
    Accepted
    Rejected
    Rejected
     
    View Code

    大意是给你三个点,绘制一个圆形区域,再输入一个点,问在不在圆内

    难点:高精度

      1 #include <cmath>
      2 #include <map>
      3 #include <vector>
      4 #include <cstdio>
      5 #include <cstdlib>
      6 #include <iostream>
      7 #include <algorithm>
      8 #include <cstring>
      9 #include <set>
     10 #include <bitset>
     11 #include <memory.h>
     12 #include <functional>
     13 #include <queue>
     14 #include <fstream>
     15 #include <ctime>
     16 #include <deque>
     17 #include <utility>
     18 #include <stack>
     19 #include <sstream>
     20 #include <list>
     21 #include <cctype> 
     22 #include <numeric>
     23 #include <iomanip>
     24 #include <assert.h>
     25 #define EPS 1e-8
     26 using namespace std;
     27 const int maxn = 550;
     28 typedef long long LL;
     29 
     30 #define MAX_L 2005 
     31 
     32 class bign
     33 {
     34 public:
     35     int len, s[MAX_L];
     36     bign();
     37     bign(const char*);
     38     bign(int);
     39     bool sign;
     40     string toStr() const;
     41     friend istream& operator>>(istream &,bign &);
     42     friend ostream& operator<<(ostream &,bign &);
     43     bign operator=(const char*);
     44     bign operator=(int);
     45     bign operator=(const string);
     46     bool operator>(const bign &) const;
     47     bool operator>=(const bign &) const;
     48     bool operator<(const bign &) const;
     49     bool operator<=(const bign &) const;
     50     bool operator==(const bign &) const;
     51     bool operator!=(const bign &) const;
     52     bign operator+(const bign &) const;
     53     bign operator++();
     54     bign operator++(int);
     55     bign operator+=(const bign&);
     56     bign operator-(const bign &) const;
     57     bign operator--();
     58     bign operator--(int);
     59     bign operator-=(const bign&);
     60     bign operator*(const bign &)const;
     61     bign operator*(const int num)const;
     62     bign operator*=(const bign&);
     63     bign operator/(const bign&)const;
     64     bign operator/=(const bign&);
     65     bign operator%(const bign&)const;
     66     bign factorial()const;
     67     bign Sqrt()const;
     68     bign pow(const bign&)const;
     69     void clean();
     70     ~bign();
     71 };
     72 #define max(a,b) a>b ? a : b
     73 #define min(a,b) a<b ? a : b
     74 
     75 bign::bign()
     76 {
     77     memset(s, 0, sizeof(s));
     78     len = 1;
     79     sign = 1;
     80 }
     81 
     82 bign::bign(const char *num)
     83 {
     84     *this = num;
     85 }
     86 
     87 bign::bign(int num)
     88 {
     89     *this = num;
     90 }
     91 
     92 string bign::toStr() const
     93 {
     94     string res;
     95     res = "";
     96     for (int i = 0; i < len; i++)
     97         res = (char)(s[i] + '0') + res;
     98     if (res == "")
     99         res = "0";
    100     if (!sign&&res != "0")
    101         res = "-" + res;
    102     return res;
    103 }
    104 
    105 istream &operator>>(istream &in, bign &num)
    106 {
    107     string str;
    108     in>>str;
    109     num=str;
    110     return in;
    111 }
    112 
    113 ostream &operator<<(ostream &out, bign &num)
    114 {
    115     out<<num.toStr();
    116     return out;
    117 }
    118 
    119 bign bign::operator=(const char *num)
    120 {
    121     memset(s, 0, sizeof(s));
    122     char a[MAX_L] = "";
    123     if (num[0] != '-')
    124         strcpy(a, num);
    125     else
    126         for (int i = 1; i < strlen(num); i++)
    127             a[i - 1] = num[i];
    128     sign = !(num[0] == '-');
    129     len = strlen(a);
    130     for (int i = 0; i < strlen(a); i++)
    131         s[i] = a[len - i - 1] - 48;
    132     return *this;
    133 }
    134 
    135 bign bign::operator=(int num)
    136 {
    137     char temp[MAX_L];
    138     sprintf(temp, "%d", num);
    139     *this = temp;
    140     return *this;
    141 }
    142 
    143 bign bign::operator=(const string num)
    144 {
    145     const char *tmp;
    146     tmp = num.c_str();
    147     *this = tmp;
    148     return *this;
    149 }
    150 
    151 bool bign::operator<(const bign &num) const
    152 {
    153     if (sign^num.sign)
    154         return num.sign;
    155     if (len != num.len)
    156         return len < num.len;
    157     for (int i = len - 1; i >= 0; i--)
    158         if (s[i] != num.s[i])
    159             return sign ? (s[i] < num.s[i]) : (!(s[i] < num.s[i]));
    160     return !sign;
    161 }
    162 
    163 bool bign::operator>(const bign&num)const
    164 {
    165     return num < *this;
    166 }
    167 
    168 bool bign::operator<=(const bign&num)const
    169 {
    170     return !(*this>num);
    171 }
    172 
    173 bool bign::operator>=(const bign&num)const
    174 {
    175     return !(*this<num);
    176 }
    177 
    178 bool bign::operator!=(const bign&num)const
    179 {
    180     return *this > num || *this < num;
    181 }
    182 
    183 bool bign::operator==(const bign&num)const
    184 {
    185     return !(num != *this);
    186 }
    187 
    188 bign bign::operator+(const bign &num) const
    189 {
    190     if (sign^num.sign)
    191     {
    192         bign tmp = sign ? num : *this;
    193         tmp.sign = 1;
    194         return sign ? *this - tmp : num - tmp;
    195     }
    196     bign result;
    197     result.len = 0;
    198     int temp = 0;
    199     for (int i = 0; temp || i < (max(len, num.len)); i++)
    200     {
    201         int t = s[i] + num.s[i] + temp;
    202         result.s[result.len++] = t % 10;
    203         temp = t / 10;
    204     }
    205     result.sign = sign;
    206     return result;
    207 }
    208 
    209 bign bign::operator++()
    210 {
    211     *this = *this + 1;
    212     return *this;
    213 }
    214 
    215 bign bign::operator++(int)
    216 {
    217     bign old = *this;
    218     ++(*this);
    219     return old;
    220 }
    221 
    222 bign bign::operator+=(const bign &num)
    223 {
    224     *this = *this + num;
    225     return *this;
    226 }
    227 
    228 bign bign::operator-(const bign &num) const
    229 {
    230     bign b=num,a=*this;
    231     if (!num.sign && !sign)
    232     {
    233         b.sign=1;
    234         a.sign=1;
    235         return b-a;
    236     }
    237     if (!b.sign)
    238     {
    239         b.sign=1;
    240         return a+b;
    241     }
    242     if (!a.sign)
    243     {
    244         a.sign=1;
    245         b=bign(0)-(a+b);
    246         return b;
    247     }
    248     if (a<b)
    249     {
    250         bign c=(b-a);
    251         c.sign=false;
    252         return c;
    253     }
    254     bign result;
    255     result.len = 0;
    256     for (int i = 0, g = 0; i < a.len; i++)
    257     {
    258         int x = a.s[i] - g;
    259         if (i < b.len) x -= b.s[i];
    260         if (x >= 0) g = 0;
    261         else
    262         {
    263             g = 1;
    264             x += 10;
    265         }
    266         result.s[result.len++] = x;
    267     }
    268     result.clean();
    269     return result;
    270 }
    271 
    272 bign bign::operator * (const bign &num)const
    273 {
    274     bign result;
    275     result.len = len + num.len;
    276 
    277     for (int i = 0; i < len; i++)
    278         for (int j = 0; j < num.len; j++)
    279             result.s[i + j] += s[i] * num.s[j];
    280 
    281     for (int i = 0; i < result.len; i++)
    282     {
    283         result.s[i + 1] += result.s[i] / 10;
    284         result.s[i] %= 10;
    285     }
    286     result.clean();
    287     result.sign = !(sign^num.sign);
    288     return result;
    289 }
    290 
    291 bign bign::operator*(const int num)const
    292 {
    293     bign x = num;
    294     bign z = *this;
    295     return x*z;
    296 }
    297 bign bign::operator*=(const bign&num)
    298 {
    299     *this = *this * num;
    300     return *this;
    301 }
    302 
    303 bign bign::operator /(const bign&num)const
    304 {
    305     bign ans;
    306     ans.len = len - num.len + 1;
    307     if (ans.len < 0)
    308     {
    309         ans.len = 1;
    310         return ans;
    311     }
    312 
    313     bign divisor = *this, divid = num;
    314     divisor.sign = divid.sign = 1;
    315     int k = ans.len - 1;
    316     int j = len - 1;
    317     while (k >= 0)
    318     {
    319         while (divisor.s[j] == 0) j--;
    320         if (k > j) k = j;
    321         char z[MAX_L];
    322         memset(z, 0, sizeof(z));
    323         for (int i = j; i >= k; i--)
    324             z[j - i] = divisor.s[i] + '0';
    325         bign dividend = z;
    326         if (dividend < divid) { k--; continue; }
    327         int key = 0;
    328         while (divid*key <= dividend) key++;
    329         key--;
    330         ans.s[k] = key;
    331         bign temp = divid*key;
    332         for (int i = 0; i < k; i++)
    333             temp = temp * 10;
    334         divisor = divisor - temp;
    335         k--;
    336     }
    337     ans.clean();
    338     ans.sign = !(sign^num.sign);
    339     return ans;
    340 }
    341 
    342 bign bign::operator/=(const bign&num)
    343 {
    344     *this = *this / num;
    345     return *this;
    346 }
    347 
    348 bign bign::operator%(const bign& num)const
    349 {
    350     bign a = *this, b = num;
    351     a.sign = b.sign = 1;
    352     bign result, temp = a / b*b;
    353     result = a - temp;
    354     result.sign = sign;
    355     return result;
    356 }
    357 
    358 bign bign::pow(const bign& num)const
    359 {
    360     bign result = 1;
    361     for (bign i = 0; i < num; i++)
    362         result = result*(*this);
    363     return result;
    364 }
    365 
    366 bign bign::factorial()const
    367 {
    368     bign result = 1;
    369     for (bign i = 1; i <= *this; i++)
    370         result *= i;
    371     return result;
    372 }
    373 
    374 void bign::clean()
    375 {
    376     if (len == 0) len++;
    377     while (len > 1 && s[len - 1] == '')
    378         len--;
    379 }
    380 
    381 bign bign::Sqrt()const
    382 {
    383     if(*this<0)return -1;
    384     if(*this<=1)return *this;
    385     bign l=0,r=*this,mid;
    386     while(r-l>1)
    387     {
    388         mid=(l+r)/2;
    389         if(mid*mid>*this)
    390             r=mid;
    391         else
    392             l=mid;
    393     }
    394     return l;
    395 }
    396 bign::~bign()
    397 {
    398 }
    399 struct point{
    400     bign x, y;
    401 };
    402 bign get_distance( point &a,  point &b)
    403 { 
    404     return ((a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y)); 
    405 }
    406 /*
    407 point get_circle_center(const point a, const point b, const point c)
    408 {
    409     point center;
    410     bign a1 = b.x - a.x;
    411     bign b1 = b.y - a.y;
    412     bign c1 = (a1 * a1 + b1 * b1) / 2.0;
    413     bign a2 = c.x - a.x;
    414     bign b2 = c.y - a.y;
    415     bign c2 = (a2 * a2 + b2 * b2) / 2.0;
    416     bign d = a1 * b2 - a2 * b1;
    417     center.x = a.x + (c1 * b2 - c2 * b1) / d;
    418     center.y = a.y + (a1 * c2 - a2 * c1) / d;
    419     return center;
    420 }
    421 */
    422 bign D;
    423 point get_circle_center( point a,  point b,  point c)
    424 {
    425     point center;
    426     bign a1 = b.x - a.x;
    427     bign b1 = b.y - a.y;
    428     bign c1 = (a1 * a1 + b1 * b1);
    429     bign a2 = c.x - a.x;
    430     bign b2 = c.y - a.y;
    431     bign c2 = (a2 * a2 + b2 * b2);
    432     bign d = a1 * b2 - a2 * b1;
    433     D = d * 2;
    434     center.x = a.x * 2 * d + (c1 * b2 - c2 * b1);
    435     center.y = a.y * 2 * d + (a1 * c2 - a2 * c1);
    436     return center;
    437 }
    438 void solve()
    439 {
    440     point a, b, c, d;
    441     //scanf("%lf%lf%lf%lf%lf%lf%lf%lf", &a.x, &a.y, &b.x, &b.y, &c.x, &c.y, &d.x, &d.y);
    442     cin >> a.x >> a.y >> b.x >> b.y >> c.x >> c.y >> d.x >> d.y;
    443     /*
    444     int ax, ay, bx, by, cx, cy, dx, dy;
    445     cin >> ax >> ay >> bx >> by >> cx >> cy >> dx >> dy;
    446     a.x = ax + 1000000000;
    447     a.y = ay + 1000000000;
    448     b.x = bx + 1000000000;
    449     b.y = by + 1000000000;
    450     c.x = cx + 1000000000;
    451     c.y = cy + 1000000000;
    452     d.x = dx + 1000000000;
    453     d.y = dy + 1000000000;
    454     */
    455     point o = get_circle_center(a, b, c);
    456     a.x = a.x * D, a.y = a.y * D, d.x = d.x * D, d.y = d.y * D;
    457     //cout << get_distance(o, a).toStr() << " " << get_distance(o, d).toStr() << endl;
    458     if (get_distance(o, a) < get_distance(o, d))
    459         puts("Accepted");
    460     else
    461         puts("Rejected");
    462 }
    463 int main()
    464 {
    465     int test;
    466     scanf("%d", &test);
    467     for (int i = 0; i < test; ++i)
    468         solve();
    469     return 0;
    470 }
    C++
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  • 原文地址:https://www.cnblogs.com/upstart/p/7543057.html
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