P4030 [Code+#2]可做题1

https://www.luogu.com.cn/record/94434358

Luogu/P4030/P4030.cpp

@@ -0,0 +1,49 @@
+#include <iostream>
+
+using std::cin;
+using std::cout;
+const char endl = '\n';
+
+const int N = 505;
+
+int n, m, t, a[N][N], s[N][N];
+bool g[N][N];
+
+int sum(int x1, int y1, int x2, int y2) {
+    return s[x2][y2] - s[x1 - 1][y2] - s[x2][y1 - 1] + s[x1 - 1][y1 - 1];
+}
+
+int main() {
+    std::ios::sync_with_stdio(false);
+    cin.tie(nullptr);
+
+    cin >> n >> m >> t;
+
+    for (int i = 1; i <= n; i++) {
+        for (int j = 1; j <= m; j++) {
+            cin >> a[i][j];
+        }
+    }
+
+    for (int i = 1; i < n; i++) {
+        for (int j = 1; j < m; j++) {
+            g[i][j] = a[i][j] + a[i + 1][j + 1] == a[i + 1][j] + a[i][j + 1];
+        }
+    }
+
+    for (int i = 1; i < n; i++) {
+        for (int j = 1; j < m; j++) {
+            s[i][j] = s[i - 1][j] + s[i][j - 1] - s[i - 1][j - 1] + g[i][j];
+        }
+    }
+
+    while (t--) {
+        int x, y, k;
+
+        cin >> x >> y >> k;
+
+        cout << (k == 1 || sum(x, y, x + k - 2, y + k - 2) == (k - 1) * (k - 1) ? 'Y' : 'N') << endl;
+    }
+
+    return 0;
+}