zoukankan      html  css  js  c++  java
  • codeforces 621B Wet Shark and Bishops

    B. Wet Shark and Bishops

     

    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.

    #include<cstdio>
    #include<cstring>
    #include<algorithm>
    using namespace std;
    typedef long long ll;
    typedef unsigned long long ull;
    const int INF=0x3f3f3f3f;
    bool G[1005][1005];
    int main()
    {
        int n;
        ll ans=0;
        scanf("%d",&n);
        for(int i=0;i<n;i++)
        {
            int x,y;
            scanf("%d%d",&x,&y);
            G[x][y]=1;
        }
        for(int i=1;i<=1000;i++)
        {
            int t=1,j=i,cnt=0;
            while(j<=1000)
                if(G[t++][j++])cnt++;
            ans+=cnt*(cnt-1)/2;
        }
        for(int i=2;i<=1000;i++)
        {
            int t=i,j=1,cnt=0;
            while(t<=1000)
                if(G[t++][j++])cnt++;
            ans+=cnt*(cnt-1)/2;
        }
        for(int i=1;i<=1000;i++)
        {
            int t=1000,j=i,cnt=0;
            while(j<=1000)
                if(G[t--][j++])cnt++;
            ans+=cnt*(cnt-1)/2;
        }
        for(int i=999;i>=1;i--)
        {
            int t=i,j=1,cnt=0;
            while(t>=1)
                if(G[t--][j++])cnt++;
            ans+=cnt*(cnt-1)/2;
        }
        printf("%I64d
    ",ans);
        return 0;
    }
  • 相关阅读:
    JS 笔记
    html笔记 横向两列布局
    jsp HTTP Status 405
    有效范围为request的bean
    jsp:session对象存储数据
    sql笔记
    StringBuffer的用法
    VB学习笔记
    html 笔记
    Linux 笔记
  • 原文地址:https://www.cnblogs.com/homura/p/5176293.html
Copyright © 2011-2022 走看看