codeforces 52B Right Triangles

B. Right Triangles

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given a n × m field consisting only of periods ('.') and asterisks ('*'). Your task is to count all right triangles with two sides parallel to the square sides, whose vertices are in the centers of '*'-cells. A right triangle is a triangle in which one angle is a right angle (that is, a 90 degree angle).

Input

The first line contains two positive integer numbers n and m (1 ≤ n, m ≤ 1000). The following n lines consist of m characters each, describing the field. Only '.' and '*' are allowed.

Output

Output a single number — total number of square triangles in the field. Please, do not use %lld specificator to read or write 64-bit integers in C++. It is preffered to use cout (also you may use %I64d).

Examples

input

Copy

2 2
**
*.

output

1

input

Copy

3 4
*..*
.**.
*.**

output

9

题意：给你一个n*m网格，求直角边和坐标轴平行且由*组成的直角三角形个数。

解法：暴力，直接算出每行每列的*个数，枚举每个*，贡献为(x[i]-1)*(y[i]-1)。

代码：

#include 
#include 
using namespace std;
int x[1005],y[1005],n,m;
char map[1005][1005];
long long ans;
int main()
{
	scanf("%d%d",&n,&m);
	for(int i=1;i<=n;i++) scanf("%s",map[i]+1);
	for(int i=1;i<=n;i++)
		for(int j=1;j<=m;j++)
			if(map[i][j]=='*')
				x[i]++,y[j]++;
	for(int i=1;i<=n;i++)
		for(int j=1;j<=m;j++)
			if(map[i][j]=='*')
				ans+=(x[i]-1)*(y[j]-1);
	printf("%I64d\n",ans);
}