Codeforces 612B. Wet Shark and Bishops

B. Wet Shark and Bishops

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Today, Wet Shark is given n bishops on a 1000 by 1000 grid. Both rows and columns of the grid are numbered from 1 to 1000. Rows are numbered from top to bottom, while columns are numbered from left to right.

Wet Shark thinks that two bishops attack each other if they share the same diagonal. Note, that this is the only criteria, so two bishops may attack each other (according to Wet Shark) even if there is another bishop located between them. Now Wet Shark wants to count the number of pairs of bishops that attack each other.

Input

The first line of the input contains n (1 ≤ n ≤ 200 000) — the number of bishops.

Each of next n lines contains two space separated integers xi and yi (1 ≤ xi, yi ≤ 1000) — the number of row and the number of column where i-th bishop is positioned. It's guaranteed that no two bishops share the same position.

Output

Output one integer — the number of pairs of bishops which attack each other.

Sample test(s)

input

5
1 1
1 5
3 3
5 1
5 5

output

6

input

3
1 1
2 3
3 5

output

0

Note

In the first sample following pairs of bishops attack each other: (1, 3), (1, 5), (2, 3), (2, 4), (3, 4) and (3, 5). Pairs (1, 2), (1, 4), (2, 5)and (4, 5) do not attack each other because they do not share the same diagonal.

 

题意：在1000×1000的棋盘上面有n个棋子“象”。如果两个棋子处在同一条对角线上面，它们就会相互攻击，即使对角线之间有其他的棋子隔开。求有几对相互攻击的棋子。

思路：每个棋子都有相应的坐标，那么就把它们的坐标转换成为对角线的形式。sign1数组记录为'\'形对角线上的棋子个数，s2数组记录'/'形对角线上的棋子个数,棋子(i,j)对应siang1[1000-(i,j)]和sign2[i+j-1];

#include<bits/stdc++.h>
using namespace std;
int sign1[3000],sign2[3000];
int main()
{
    int i,n,x,y,d1,d2;
    scanf("%d",&n);
    memset(sign1,0,sizeof(sign1));
    memset(sign2,0,sizeof(sign2));
    for(i=0; i<n; i++)
    {
        scanf("%d%d",&x,&y);
        d1=x-y;
        sign1[1000-d1]++;
        d2=x+y;
        sign2[d2-1]++;
    }
    __int64 ans=0;
    for(i=0; i<3000;i++)
    {
        ans+=sign1[i]*(sign1[i]-1)/2;
        ans+=sign2[i]*(sign2[i]-1)/2;
    }
    cout<<ans<<endl;
    return 0;
}