Symmetry（對稱軸存在問題），symmetry對稱軸

                    Symmetry

Description

The figure shown on the left is left-right symmetric as it is possible to fold the sheet of paper along a vertical line, drawn as a dashed line, and to cut the figure into two identical halves. The figure on the right is not left-right symmetric as it is impossible to find such a vertical line.

Write a program that determines whether a figure, drawn with dots, is left-right symmetric or not. The dots are all distinct.

Input 

The input consists of T test cases. The number of test cases T is given in the first line of the input file. The first line of each test case contains an integer N , where N ( 1N1, 000) is the number of dots in a figure. Each of the following N lines contains the x-coordinate and y-coordinate of a dot. Both x-coordinates and y-coordinates are integers between -10,000 and 10,000, both inclusive.

Output 

Print exactly one line for each test case. The line should contain `YES' if the figure is left-right symmetric. and `NO', otherwise.

The following shows sample input and output for three test cases.

Sample Input 

3                                            
5                                            
-2 5                                         
0 0 
6 5 
4 0 
2 3 
4 
2 3 
0 4 
4 0 
0 0 
4 
5 14 
6 10
5 10 
6 14

Sample Output 

YES 
NO 
YES

題意；在坐標系中給出的點中，尋找一個豎直的對稱軸，讓這些點關於它對稱，如果它存在，輸出YES，否則輸出NO

看了很多人的博客，大多數此題用STL解決，了解它會更容易，但是我不太會，所以還是介紹的是一種自己好理解的避免出現精度問題的方法（坐標*2）

AC代碼：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
using namespace std;
const double eps=1e-5;//1e-5:浮點數,在計算機中這麼表示,在數學中是科學計數法.1e-5的意思就是1乘以10的負5次冪。就是0.000001

int x[1010],y[1010];
int main()
{
　　int T;
　　cin>>T;
　　while(T--)
{
　　int n,i,j;;
　　cin>>n;

　　double sum=0;

 for(i=1; i<=n; i++)
{
　　cin>>x[i]>>y[i];
　　sum+=x[i];
}
　　sum/=n;

for(i=1; i<=n; i++)
{
　　for(j=1; j<=n; j++)
{
　　if(abs(2*sum-x[i]-x[j])<eps&&abs(y[i]-y[j])<eps)
break;
}
　　if(j>n) break;
}
　　if(i>n)
cout<<"YES"<<endl;
　　else
cout<<"NO"<<endl;
}
}