CF#341-B-Wet Shark and Bishops-水题

http://codeforces.com/contest/621/problem/B

给你一个1000*1000的矩阵，上的n个点，

计算每个对角线上有多少个点，如果一个对角线有n个点，代表有n个牧师，会互相打架，计算打架总数

直接统计每条对角线多少个点，然后 ....n个人，有n*(n-1)/2场架，GG

//注意主对角线要+n，否则下标为负数

#include <cstdio>#include <cmath>#include <cstring>#include <string>#include <algorithm>#include <iostream>#include <queue>#include <map>#include <set>#include <vector>#include<stack>using namespace std; __int64 L[2005*2];//副对角线元素之和 __int64 R[2005*2]; //主对角线元素之和   int main(){ int n;int i,x,y;scanf("%d",&n); for (i=1;i<=n;i++){ scanf("%d %d",&x,&y);  L[x+y]++;  R[x-y+1000]++;  //注意}__int64 sum=0;for (i=0;i<=2000;i++){if (L[i]) sum+= L[i]*(L[i]-1)/2;if (R[i]) sum+= R[i]*(R[i]-1)/2;}printf("%I64d\n",sum);return 0;}