zoukankan      html  css  js  c++  java
  • Codeforces Round #343 (Div. 2) A

    A. Far Relative’s Birthday Cake
    time limit per test
    1 second
    memory limit per test
    256 megabytes
    input
    standard input
    output
    standard output

    Door's family is going celebrate Famil Doors's birthday party. They love Famil Door so they are planning to make his birthday cake weird!

    The cake is a n × n square consisting of equal squares with side length 1. Each square is either empty or consists of a single chocolate. They bought the cake and randomly started to put the chocolates on the cake. The value of Famil Door's happiness will be equal to the number of pairs of cells with chocolates that are in the same row or in the same column of the cake. Famil Doors's family is wondering what is the amount of happiness of Famil going to be?

    Please, note that any pair can be counted no more than once, as two different cells can't share both the same row and the same column.

    Input

    In the first line of the input, you are given a single integer n (1 ≤ n ≤ 100) — the length of the side of the cake.

    Then follow n lines, each containing n characters. Empty cells are denoted with '.', while cells that contain chocolates are denoted by 'C'.

    Output

    Print the value of Famil Door's happiness, i.e. the number of pairs of chocolate pieces that share the same row or the same column.

    Examples
    Input
    3
    .CC
    C..
    C.C
    Output
    4
    Input
    4
    CC..
    C..C
    .CC.
    .CC.
    Output
    9
    Note

    If we number rows from top to bottom and columns from left to right, then, pieces that share the same row in the first sample are:

    1. (1, 2) and (1, 3)
    2. (3, 1) and (3, 3)
    Pieces that share the same column are:
    1. (2, 1) and (3, 1)
    2. (1, 3) and (3, 3)

    开学了 额

    题意: n*n矩阵 每行每列 C两两配对 共有多少种

    题解: 遍历一遍 记录每行每列C的个数 求组合数 相加

    #include<iostream>
    #include<cstring>
    #include<map>
    #include<cstdio>
    using namespace std;
    int n;
    char a[105][105];
    map<int,int>m;
    map<int,int>p;
    int main()
    {
        scanf("%d",&n);
        getchar();
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=n;j++)
            {
                scanf("%c",&a[i][j]);
                if(a[i][j]=='C')
                {
                    m[i]=m[i]+1;
                    p[j]=p[j]+1;
                }
            }
            getchar();
        }
        int sum=0;
        for(int i=1;i<=n;i++)
        {
            sum=sum+m[i]*(m[i]-1)/2+p[i]*(p[i]-1)/2;
        }
        printf("%d
    ",sum);
    

      

  • 相关阅读:
    在ubuntu下关闭笔记本触摸板
    (转载)实用小命令 windows下查看端口占用情况
    (转载)JBoss 4.2.3下部署EJB 3.0碰到的local和remote问题
    Windows下通过xmanager远程桌面控制Linux(转)
    SQL Server 事务日志的问题
    回归
    用友软件工程IT应用研究院
    关于Oracle数据库的死锁(转书摘)
    关于企业级Web2.0的一点想法
    关注Java的开源项目(中文版)
  • 原文地址:https://www.cnblogs.com/hsd-/p/5232306.html
Copyright © 2011-2022 走看看