Flea Circus（Project Euler 213）

original version

hackerrank programming version

题目大意是N*N的格子，每个格子一开始有1个跳蚤，每过单位时间跳蚤会等概率向四周跳，问M秒后空格子的期望个数。

题解：

对于每个跳蚤暴力模拟每一秒，算出M秒后它到各个格子的概率最后统计就好了，hackerrank上的版本需要常数优化，比如根据对称性只考虑左上四分之一块矩形.

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

 typedef long long LL;
 const int mod = 1e9 + ;
 #define MAXN 42
 #define MP make_pair
 const int INF = 1e9 + ;

 int n, m;
 int s[MAXN][MAXN], tmp[MAXN][MAXN], cnt[MAXN][MAXN];
 LL p[MAXN][MAXN];
 int dx[] = {, , -, };
 int dy[] = {, -, , };
 int inv[];

 int power(int x, int p)
 {
     int res = ;
     for (; p; p >>= )
     {
         if (p & ) res = 1LL * res * x % mod;
         x = 1LL * x * x % mod;
     }
     return res;
 }

 void work(int sx, int sy)
 {
     for (int i = ; i <= n; ++i)
         for (int j = ; j <= n; ++j)
             s[i][j] = ;
     s[sx][sy] = ;

     int x, y;
     for (int _m = ; _m <= m; ++_m)
     {
         for (int i = ; i <= n; ++i)
             for (int j = ; j <= n; ++j)
                 tmp[i][j] = ;
         for (int i = ; i <= n; ++i)
         {
             for (int j = ; j <= n; ++j)
             {
                 for (int d = ; d < ; ++d)
                 {
                     x = i + dx[d];
                     y = j + dy[d];
                     (tmp[x][y] += 1LL * cnt[i][j] * s[i][j] % mod) %= mod;
                 }
             }
         }
         for (int i = ; i <= n; ++i)
             for (int j = ; j <= n; ++j)
                 s[i][j] = tmp[i][j];
     }
     int t;
     for (int i = ; i <= n; ++i)
     {
         for (int j = ; j <= n; ++j)
         {
             t =  - s[i][j] + mod;
             (p[i][j] *= t) %= mod;
             (p[n +  - i][j] *= t) %= mod;
             (p[i][n +  - j] *= t) %= mod;
             (p[n +  - i][n +  - j] *= t) %= mod;
         }
     }
 }

 int main()
 {
     //freopen("in.txt", "r", stdin);
     //freopen("out.txt", "w", stdout);

     inv[] = ;
     inv[] = power(, mod - );
     inv[] = power(, mod - );
     inv[] = power(, mod - );

     int T, x, y;
     scanf("%d", &T);
     while (T--)
     {
         scanf("%d %d", &n, &m);
         for (int i = ; i <= n; ++i)
         {
             for (int j = ; j <= n; ++j)
             {
                 cnt[i][j] = ;
                 for (int d = ; d < ; ++d)
                 {
                     x = i + dx[d];
                     y = j + dy[d];
                     if (x >=  && x <= n && y >=  && y <= n)
                         ++cnt[i][j];
                 }
                 cnt[i][j] = inv[cnt[i][j]];
             }

         }
         for (int i = ; i <= n; ++i)
             for (int j = ; j <= n; ++j)
                 p[i][j] = ;
         LL ans = ;
         for (int i = ; i <= n / ; ++i)
             for (int j = ; j <= n / ; ++j)
                 work(i, j);
         for (int i = ; i <= n; ++i)
             for (int j = ; j <= n; ++j)
                 ans += p[i][j];
         ans %= mod;
         cout << ans << endl;
      }

     return ;
 }