Codeforces Round #341 (Div. 2) B. Wet Shark and Bishops （正副对角线规律）

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：1个1000*1000的地图，上面有一些点，如果两个点在一个对角线上，那么他们就可以互相攻击，求共有多少对这样的点

思路：正副对角线的规律，正对角线上的点i-j相同，副对角线上的点i+j相同，这样这道题就很好求了，一个数组保存

各个对角线的规律值，因为i-j会有负数，所以扩大数组，i+j极限值为2000,i-j的最小值为-999，所以说数组极限值要大于2999，最后发现一条线上的对数为一个等差数列，求和就好了。。

交了两发就睡着了，真的是日了狗了。。。后面题都没看。。。

ac代码：

#include<stdio.h>
#include<math.h>
#include<string.h>
#include<stack>
#include<set>
#include<queue>
#include<vector>
#include<iostream>
#include<algorithm>
#define MAXN 100010#define LL long long
#define ll __int64
#define INF 0x7fffffff
#define mem(x) memset(x,0,sizeof(x))
#define PI acos(-1)
#define eps 1e-10using namespace std;
int gcd(int a,int b){return b?gcd(b,a%b):a;}
LL powmod(LL a,LL b,LL MOD){LL ans=1;while(b){if(b%2)ans=ans*a%MOD;a=a*a%MOD;b/=2;}return ans;}
//head
int cnt[MAXN];
int main()
{
int n,i,j,a,b;
while(scanf("%d",&n)!=EOF)
{
mem(cnt);
for(i=0;i<n;i++)
{
scanf("%d%d",&a,&b);
cnt[a+b]++;
cnt[a-b+3000]++;
}
ll ans=0;
for(i=0;i<=4040;i++)
{
if(cnt[i])
{
ll k=(cnt[i]-1)*(cnt[i]-1)+cnt[i]-1;
ans+=k/2;
}
}
printf("%I64d\n",ans);
}
return 0;
}