前言
今天又是爆零的一天。
被同学坑了,还以为四边形的点是按任意序给定的,然后打了一个特别复杂的矩形判断QAQ。
题意简述
按顺序给定一个四边形,求有多少个点在这个四边形的对称轴上。
题解
分情况讨论:
正方形有8个点。
菱形和矩形有4个点。
筝形和等腰梯形有2个点。
剩余的有0个点。
代码
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <map>
using namespace std;
namespace fast_IO{
const int IN_LEN = 10000000, OUT_LEN = 10000000;
char ibuf[IN_LEN], obuf[OUT_LEN], *ih = ibuf + IN_LEN, *oh = obuf, *lastin = ibuf + IN_LEN, *lastout = obuf + OUT_LEN - 1;
inline char getchar_(){return (ih == lastin) && (lastin = (ih = ibuf) + fread(ibuf, 1, IN_LEN, stdin), ih == lastin) ? EOF : *ih++;}
inline void putchar_(const char x){if(oh == lastout) fwrite(obuf, 1, oh - obuf, stdout), oh = obuf; *oh ++= x;}
inline void flush(){fwrite(obuf, 1, oh - obuf, stdout);}
int read(){
int x = 0; int zf = 1; char ch = ' ';
while (ch != '-' && (ch < '0' || ch > '9')) ch = getchar_();
if (ch == '-') zf = -1, ch = getchar_();
while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar_(); return x * zf;
}
void write(int x){
if (x < 0) putchar_('-'), x = -x;
if (x > 9) write(x / 10);
putchar_(x % 10 + '0');
}
}
using namespace fast_IO;
#define x first
#define y second
#define Pos pair<double, double>
Pos pos[4];
const double EPS = 1e-8;
inline bool eql(double a, double b){
return (abs(a - b) <= EPS);
}
inline double getDis(Pos a, Pos b){
return (a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y);
}
bool ggdl(int pos1, int pos2, int pos3){
double dis1 = getDis(pos[pos1], pos[pos2]);
double dis2 = getDis(pos[pos2], pos[pos3]);
double dis3 = getDis(pos[pos1], pos[pos3]);
return (eql(dis1 + dis2, dis3) || eql(dis1 + dis3, dis2) || eql(dis2 + dis3, dis1));
}
bool isJx(){
double x_c = (pos[0].x + pos[1].x + pos[2].x + pos[3].x) / 4.0;
double y_c = (pos[0].y + pos[1].y + pos[2].y + pos[3].y) / 4.0;
double d1 = getDis(pos[0], make_pair(x_c, y_c));
double d2 = getDis(pos[1], make_pair(x_c, y_c));
double d3 = getDis(pos[2], make_pair(x_c, y_c));
double d4 = getDis(pos[3], make_pair(x_c, y_c));
return (d1 == d2 && d1 == d3 && d1 == d4);
}
bool isLx(){
double s12 = getDis(pos[0], pos[1]);
double s14 = getDis(pos[0], pos[3]);
double s23 = getDis(pos[1], pos[2]);
double s34 = getDis(pos[2], pos[3]);
return (s12 == s14 && s14 == s23 && s23 == s34);
}
bool isZs(){
bool zs1 = eql(getDis(pos[0], pos[1]), getDis(pos[1], pos[2])) && eql(getDis(pos[0], pos[3]), getDis(pos[3], pos[2]));
bool zs2 = eql(getDis(pos[0], pos[1]), getDis(pos[0], pos[3])) && eql(getDis(pos[2], pos[1]), getDis(pos[2], pos[3]));
return (zs1 || zs2);
}
bool tx(){
bool x = eql(getDis(pos[0], pos[1]), getDis(pos[2], pos[3]));
x = x || (eql(getDis(pos[1], pos[2]), getDis(pos[0], pos[3])));
return (eql(getDis(pos[0], pos[2]), getDis(pos[1], pos[3])) && x);
}
int main(){
for (int i = 0; i < 4; ++i) pos[i].x = read(), pos[i].y = read();
bool jx = isJx(), lx = isLx();
if (jx && lx)
puts("8");
else if (jx || lx)
puts("4");
else{
if (isZs() || tx())
puts("2");
else
puts("0");
}
return 0;
}