2 条题解
-
1
using namespace std; int main() { ios::sync_with_stdio(false); cin.tie(nullptr); int n, k; cin >> n >> k; vector<vector<char>> lit(n, vector<char>(n, 0)); for (int i = 0; i < k; ++i) { int x, y; cin >> x >> y; --x; --y; for (int dx = -1; dx <= 1; ++dx) for (int dy = -1; dy <= 1; ++dy) { int nx = x + dx, ny = y + dy; if (0 <= nx && nx < n && 0 <= ny && ny < n) lit[nx][ny] = 1; } } long long cnt = 0; for (int i = 0; i < n; ++i) cnt += count(lit[i].begin(), lit[i].end(), 1); cout << (long long)n * n - cnt << '\n'; return 0; }
- 1
信息
- ID
- 3105
- 时间
- 1000ms
- 内存
- 256MiB
- 难度
- 2
- 标签
- 递交数
- 37
- 已通过
- 15
- 上传者