思路:求出当前两个位置上放上横条或者数条的概率,答案即为 2 w ∗ P 2^w*P 2w∗P;
#include <iostream>
using namespace std;
typedef long long LL;
const int N = 3e5 + 10;
const int mod = 998244353;
LL qmi(int a, int b, int mod)
{
LL res = 1;
while (b)
{
if (b & 1) res = (LL)res * a % mod;
a = (LL)a * a % mod;
b >>= 1;
}
return res;
}
int inv(int x)
{
return qmi(x, mod - 2, mod);
}
int divide(int x)
{
if (x % 2 == 0) return 1 * inv(qmi(2, x, mod));
else return -1 * inv(qmi(2, x, mod));
}
LL f[N], sum[N];
string s[N];
int main()
{
for (int i = 2; i < N; ++i)
{
f[i] = f[i - 1] + divide(i);
sum[i] = (sum[i - 1] + f[i]) % mod;
}
int n, m;
cin >> n >> m;
for (int i = 0; i < n; ++i) cin >> s[i];
LL res = 0, cnt = 0;
for (int i = 0; i < n; ++i)
for (int j = 0; j < m; ++j)
if (s[i][j] == 'o') cnt++;
for (int i = 0; i < n; ++i)
for (int j = 0; j < m; ++j)
{
if (s[i][j] == 'o')
{
int tmp = 1;
while (j + 1 < m && s[i][j + 1] == 'o') j++, tmp++;
res = (res + sum[tmp]) % mod;
}
}
for (int i = 0; i < m; ++i)
for (int j = 0; j < n; ++j)
{
if (s[j][i] == 'o')
{
int tmp = 1;
while (j + 1 < n && s[j + 1][i] == 'o') j++, tmp++;
res = (res + sum[tmp]) % mod;
}
}
cout << (res * qmi(2, cnt, mod) % mod + mod) % mod << endl;
return 0;
}