Q - Euclid in Manhattan（欧几里德距离==曼哈顿距离?）

Desciption

Consider a set of n points in a 2-D plane with integer coordinates. There are various ways of measuring distance between two points such as Euclidean , Manhattan , Chebyshev distance. These distances have important application , one of which is chess.

Consider that the ith point is located at (xi , yi). We want to find the number of pairs(i, j) such that the Euclidean distance between the points i and j is equal to the Manhattan distance between the same two points, i.e. Euclidean distance(i, j) = Manhattan distance(i, j).

√((xi − xj )^2 + (yi − yj )^2) - is called Euclidean distance

| xi − xj | + | yi − yj | - is called Manhattan distance

Note - All the n points given are considered different, even if they share the same coordinates.

Input

First line contains n, number of points in the plane Each of the following n lines contains two integers xi , yi

Output

Print the total number of such pairs.

Example

Input:
3
1 1
7 5
1 5
Output:
2 

Input:
6
0 0
0 1
0 2
-1 1
0 1
1 1
Output:
11
解题思路：注意判断两个小数是否相等，一般采用作差法。如果两个小数的差值小于一个很小的精度，则视这两个小数相等。这题没给出n的范围，后台测试数据比较小，暴力O(n^2)水过。
AC代码：

 #include<iostream>
 #include<algorithm>
 #include<cmath>
 using namespace std;
 const int maxn=1e6+;
 int n,m=;double x[maxn],y[maxn];
 const double eps=1e-;
 double Eulc(double a1,double b1,double a2,double b2){
     return sqrt((a1-a2)*(a1-a2)+(b1-b2)*(b1-b2));
 }
 double Manh(double a1,double b1,double a2,double b2){
     return abs(a1-a2)+abs(b1-b2);
 }
 int main(){
     cin>>n;
     for(int i=;i<n;++i)cin>>x[i]>>y[i];
     for(int i=;i<n-;++i){
         for(int j=i+;j<n;++j){
             if(abs(Eulc(x[i],y[i],x[j],y[j])-Manh(x[i],y[i],x[j],y[j]))<eps)m++;
         }
     }
     cout<<m<<endl;
     return ;
 }