60pts

代码:

#include<iostream>
using namespace std;
typedef long long ll;
int n,m,ans,sum;
ll a[51],b[51];
string s;
bool add(ll X)
{
    for(int i = 50;i >= 0;i--)
    {
        if(X & (1LL << i))
        {
            if(b[i])
                X ^= b[i];
            else
            {
                b[i] = X;
                return true;
            }
        }
    }
    return false;
}
int main()
{
    cin >> n >> m;
    for(int i = 1;i <= m;i++)
    {
        sum = 0;
        cin >> s;
        for(int j = 0;j < s.size();j++)
            if(s[j] == 'O')
                sum ^= (1LL << j);
        ans += add(sum);
    }
    cout << (1LL << ans) % 2008;
    return 0;
}