'commit'

TangDou/LanQiaoBei/LanQiao14STEMA20230212/8.cpp

@@ -12,7 +12,6 @@ int main() {
         cnt += 1;
     else if (n % 3 == 2)
         cnt += 2;
-
     cout << cnt << endl;
     return 0;
 }

TangDou/LanQiaoBei/LanQiao14STEMA20230212/9.cpp

@@ -0,0 +1,35 @@
+#include <bits/stdc++.h>
+using namespace std;
+const int N = 110;
+int a[N][N], s[N][N];
+/*
+4 5
+1 1 0 0 0
+1 0 1 0 0
+0 0 0 1 1
+0 0 0 1 0
+*/
+int main() {
+    int n, m;
+    cin >> n >> m;
+    for (int i = 1; i <= n; i++)
+        for (int j = 1; j <= m; j++)
+            cin >> a[i][j];
+
+    // 计算二维前缀和
+    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] + a[i][j];
+
+    int res = -1;
+    for (int x1 = 1; x1 <= n; x1++)
+        for (int y1 = 1; y1 <= m; y1++)
+            for (int x2 = x1; x2 <= n; x2++)
+                for (int y2 = y1; y2 <= m; y2++) { // 四层循环枚举每个矩形的左上角和右下角
+                    if (s[x2][y2] - s[x1 - 1][y2] - s[x2][y1 - 1] + s[x1 - 1][y1 - 1] == 0)
+                        res = max(res, (x2 - x1 + 1) * (y2 - y1 + 1));
+                }
+
+    cout << res << endl;
+    return 0;
+}