B. Wet Shark and Bishops(思维)

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 to1000. 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 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.

题解：

在同一斜线上的象可以相互攻击，问可以攻击的对数；由于x+y相等的象在同一/上，相减相等的在同一上，计算相等的数目，对数就是n*(n-1)/2；

由于相减可能为负，加一个整数就可以了；

代码：

#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
const int INF=0x3f3f3f3f;
#define mem(x,y) memset(x,y,sizeof(x))
#define SI(x) scanf("%d",&x)
#define SL(x) scanf("%I64d",&x)
#define PI(x) printf("%d",x)
#define PL(x) printf("%I64d",x)
#define P_ printf(" ")
typedef __int64 LL;
const int MAXN=4040;
int m[MAXN];
int main(){
	int n;
	while(~SI(n)){
		mem(m,0);
		int a,b;
		for(int i=0;i<n;i++){
			SI(a);SI(b);
			m[a+b]++;
			m[a-b+3020]++;
		}
		LL ans=0;
		for(int i=0;i<MAXN;i++){
			if(m[i]){
			//	printf("%d %d
",i,m[i]);
				ans+=(m[i]-1)*m[i]/2;
			}
		}
		printf("%I64d
",ans);
	}
	return 0;
}