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Codeforces Round #249 (Div. 2) (ABCD题解)(三)
& dp[i - k][j][6] >= k) ans1 ++; if(dp[i][j][0] >= k && dp[i][j][1] >= k && i - k >= 1 && dp[i - k][j][2] >= k) ans1 ++; if(dp[i][j][2] >= k && dp[i][j][1] >= k && j + k <= m && dp[i][j + k][0] >= k) ans1 ++; if(dp[i][j][2] >= k && dp[i][j][3] >= k && j + k <= m && dp[i][j + k][4] >= k) ans1 ++; if(dp[i][j][4] >= k && dp[i][j][3] >= k && i + k <= n && dp[i + k][j][2] >= k) ans1 ++; if(dp[i][j][4] >= k && dp[i][j][5] >= k && i + k <= n && dp[i + k][j][6] >= k) ans1 ++; if(dp[i][j][6] >= k && dp[i][j][5] >= k && j - k >= 1 && dp[i][j - k][4] >= k) ans1 ++; if(dp[i][j][6] >= k && dp[i][j][7] >= k && j - k >= 1 && dp[i][j - k][0] >= k) ans1 ++; } } } ans1 /= 2; int ans2 = 0; for(int i = 1; i <= n; i++) { for(int j = 1; j <= m; j++) { for(int k = 1; k <= max(n, m); k++) { if(dp[i][j][1] >= k && dp[i][j][3] >= k && i - k >= 1 && j + k <= m && dp[i - k][j + k][4] >= 2 * k) ans2 ++; if(dp[i][j][5] >= k && dp[i][j][3] >= k && i + k <= n && j - k >= 1 && dp[i + k][j - k][2] >= 2 * k) ans2 ++; if(dp[i][j][5] >= k && dp[i][j][7] >= k && i - k >= 1 && j - k >= 1 && dp[i - k][j - k][4] >= 2 * k) ans2 ++; if(dp[i][j][1] >= k && dp[i][j][7] >= k && i - k >= 1 && j - k >= 1 && dp[i - k][j - k][2] >= 2 * k) ans2 ++; } } } printf(%d , ans1 + ans2); }