Codeforces Round #343 (Div. 2) A. Far Relative’s Birthday Cake 水题

A. Far Relative’s Birthday Cake

题目连接：

http://www.codeforces.com/contest/629/problem/A

Description

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.

Sample Input

4
CC..
C..C
.CC.
.CC.

Sample Output

9

Hint

题意

给你一个n*n的矩阵，然后矩阵分钟的C会和在他同一行、同一列的C发生关系

然后问你一共发生多少次关系（嘿嘿嘿

题解：

水题啦

统计每一行有多少个，每一列有多少个，然后n*(n-1)/2就好了

代码

#include<bits/stdc++.h>
using namespace std;

char s[120][120];

int main()
{
    int n;
    scanf("%d",&n);
    for(int i=0;i<n;i++)
        scanf("%s",s[i]);
    int tot = 0;
    long long ans = 0;
    for(int i=0;i<n;i++)
    {
        int tot = 0;
        for(int j=0;j<n;j++)
            if(s[i][j]=='C')
                tot++;
        ans += tot*(tot-1)/2;
        tot = 0;
        for(int j=0;j<n;j++)
            if(s[j][i]=='C')
                tot++;
        ans+=tot*(tot-1)/2;
    }
    cout<<ans<<endl;
}