B.Wet Shark and Bishops--- Codeforces Round #341 (Div. 2)

Wet Shark and Bishops

题目链接https://codeforces.com/problemset/problem/621/B
time limit per test 2 seconds
memory limit per test 256 megabytes

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 x_i and y_i(1<=x_i,y_i<=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

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

Examples

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.

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题目大意：给你n个坐标，规定对角线会冲突，问你有多少个冲突对。
emmm，这道题让人不禁想起了八皇后问题。。。我们知道与主对角线平行的相同对角线的|x-y|是一样的，其余对角线则是x+y一样。那么我们很容易算出每个对角线起冲突的个数：

vis[x+y][0]++;
vis[x-y][1]++; 

其中1保存主对角线平行的冲突个数，0保存其他的。当然，x-y有可能是负数，所以我们要给他加上一个临界值：vis[x-y+p][1]++。那么冲突的对数就是：

(vis[i][0])*(vis[i][0]-1)/2;

即从vis[i][0]一直加到1的总和。
那么接下来的代码就很好写了：

#include <cstdio>
#define ll long long
const int mac=4000;
const int p=1100;
int vis[mac][3];
int main()
{
	int n,x,y;
	ll sum=0;
	scanf ("%d",&n);
	for (int i=1; i<=n; i++){
		scanf ("%d%d",&x,&y);
		vis[x+y+p][0]++;
		vis[x-y+p][1]++; 
	}
	for (int i=1; i<=mac; i++){
		if (vis[i][0]>1){
			sum+=(ll)(vis[i][0])*(vis[i][0]-1)/2;
		}
		if (vis[i][1]>1){
			sum+=(ll)(vis[i][1])*(vis[i][1]-1)/2;
		}
	}
	printf ("%lld\n",sum);
	return 0;
}