zoukankan      html  css  js  c++  java
  • Codeforces Round #246 (Div. 2) —B. Football Kit

    版权声明:本文为博主原创文章,未经博主同意不得转载。 https://blog.csdn.net/u013497151/article/details/25934129
    B. Football Kit
    time limit per test
    1 second
    memory limit per test
    256 megabytes
    input
    standard input
    output
    standard output

    Consider a football tournament where n teams participate. Each team has two football kits: for home games, and for away games. The kit for home games of the i-th team has color xi and the kit for away games of this team has color yi (xi ≠ yi).

    In the tournament, each team plays exactly one home game and exactly one away game with each other team (n(n - 1) games in total). The team, that plays the home game, traditionally plays in its home kit. The team that plays an away game plays in its away kit. However, if two teams has the kits of the same color, they cannot be distinguished. In this case the away team plays in its home kit.

    Calculate how many games in the described tournament each team plays in its home kit and how many games it plays in its away kit.

    Input

    The first line contains a single integer n (2 ≤ n ≤ 105) — the number of teams. Next n lines contain the description of the teams. The i-th line contains two space-separated numbers xiyi (1 ≤ xi, yi ≤ 105xi ≠ yi) — the color numbers for the home and away kits of the i-th team.

    Output

    For each team, print on a single line two space-separated integers — the number of games this team is going to play in home and away kits, correspondingly. Print the answers for the teams in the order they appeared in the input.

    Sample test(s)
    input
    2
    1 2
    2 1
    
    output
    2 0
    2 0
    
    input
    3
    1 2
    2 1
    1 3
    
    output
    3 1
    4 0
    2 2
    

    题意:打印当前队 穿主场衣服作战的次数 和  穿客场衣服作战的次数

    思路:简单的哈希思想+规律  就可以AC。仅仅是 题意不是非常好理解

    规律是 n个队  每一个队会进行 n-1次 比赛  主场服使用次数 + 客场服使用次数  = 2*(n-1)
    主场服使用次数  =  n-1 + 当前队客场服与其它队 主场服同样的次数

    #include <stdio.h>
    #include <stdlib.h>
    #include <string.h>
    
    int a[100010];
    int b[100010],c[100010];
    
    int main()
    {
        int n,i;
        scanf("%d",&n);
        memset(a,0,sizeof(a));
        for(i = 0; i<n; i++)
        {
            scanf("%d%d",&b[i],&c[i]);
            a[b[i]]++;
        }
        for(i = 0; i<n; i++)
        {
            printf("%d %d
    ",n-1+a[c[i]],2*(n-1)-(n-1+a[c[i]]));
        }
        return 0;
    }


  • 相关阅读:
    12 【结构型】 浅谈享元模式的理解与使用
    11【结构型】浅谈 外观模式 理解与使用~
    【Maven基础入门】02 了解POM文件构建
    【Maven基础入门】01 Maven的安装与环境变量的配置
    02【创建型】原型模式
    01【创建型】单例模式
    10 浅谈 装饰器模式的理解与使用
    Java JDK1.8源码学习之路 2 String
    Java JDK1.8源码学习之路 1 Object
    ApachShiro 一个系统 两套验证方法-(后台管理员登录、前台App用户登录)同一接口实现、源码分析
  • 原文地址:https://www.cnblogs.com/mqxnongmin/p/10657068.html
Copyright © 2011-2022 走看看