'commit'

TangDou/LanQiaoBei/LanQiao14STEMA202302/1.cpp

@@ -0,0 +1,9 @@
+#include <bits/stdc++.h>
+
+using namespace std;
+int main() {
+    int n;
+    cin >> n;
+    printf("%d\n", n - 1 + n + 1);
+    return 0;
+}

TangDou/LanQiaoBei/LanQiao14STEMA202302/2.cpp

@@ -0,0 +1,21 @@
+#include <bits/stdc++.h>
+
+using namespace std;
+int n, m;
+int cnt;
+bool check(int x) {
+    while (x) {
+        int r = x % 10;
+        if (r % 2 == 0) return false;
+        x /= 10;
+    }
+    return true;
+}
+int main() {
+    cin >> n >> m;
+    for (int i = n; i <= m; i++)
+        if (check(i)) cnt++;
+
+    printf("%d\n", cnt);
+    return 0;
+}

TangDou/LanQiaoBei/LanQiao14STEMA202302/3.cpp

@@ -0,0 +1,33 @@
+#include <bits/stdc++.h>
+using namespace std;
+int n;
+const int INF = 0x3f3f3f3f;
+/*
+手推4
+0 0 0 0
+1 1 1 0
+0 0 1 1
+0 1 0 0
+1 1 1 1
+共4步
+
+手推5
+0 0 0 0 0
+1 1 1 0 0
+0 0 1 1 0
+1 1 1 1 1
+共3步
+*/
+void solve(int n) {
+    if (n % 3 == 0)
+        cout << n / 3 << " ";
+    else if (n % 3 == 1)
+        cout << n / 3 - 1 + 4 << " "; // 4是4张手推结果
+    else
+        cout << n / 3 - 1 + 3 << " "; // 3是5张手推结果
+}
+
+int main() {
+    for (int i = 1; i <= 20; i++) solve(i);
+    return 0;
+}

TangDou/LanQiaoBei/LanQiao14STEMA202302/4.cpp

@@ -0,0 +1,35 @@
+#include <bits/stdc++.h>
+
+using namespace std;
+const int N = 110;
+int n, m;
+int a[N][N], s[N][N];
+int res;
+/*
+4 5
+1 1 0 0 0
+1 0 1 0 0
+0 0 0 1 1
+0 0 0 1 0
+
+答案：6
+*/
+int main() {
+    cin >> n >> m;
+    for (int i = 1; i <= n; i++)
+        for (int j = 1; j <= m; j++) {
+            cin >> a[i][j];
+            s[i][j] = s[i - 1][j] + s[i][j - 1] - s[i - 1][j - 1] + a[i][j];
+        }
+
+    for (int x1 = 1; x1 <= n; x1++)
+        for (int y1 = 1; y1 <= m; y1++)
+            for (int x2 = x1 + 1; x2 <= n; x2++)
+                for (int y2 = y1 + 1; y2 <= m; y2++) {
+                    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));
+                }
+
+    printf("%d\n", res);
+    return 0;
+}

TangDou/LanQiaoBei/LanQiao14STEMA202302/TD.cpp

@@ -0,0 +1,26 @@
+#include <bits/stdc++.h>
+
+using namespace std;
+const int N = 110;
+int n, m;
+int a[N][N], s[N][N];
+int res;
+int main() {
+    cin >> n >> m;
+    for (int i = 1; i <= n; i++)
+        for (int j = 1; j <= m; j++) {
+            cin >> a[i][j];
+            a[i][j] = !a[i][j];
+            s[i][j] = s[i - 1][j] + s[i][j - 1] - s[i - 1][j - 1] - a[i][j];
+        }
+    for (int x1 = 1; x1 <= n; x1++)
+        for (int y1 = 1; y1 <= m; y1++)
+            for (int x2 = x1 + 1; x2 <= n; x2++)
+                for (int y2 = y1 + 1; y2 <= m; y2++) {
+                    int sum = s[x2][y2] - s[x1 - 1][y2] - s[x2][y1 - 1] + s[x1 - 1][y1 - 1];
+                    if ((x2 - x1 + 1) * (y2 - y1 + 1) == sum)
+                        res = max(res, sum);
+                }
+    printf("%d\n", res);
+    return 0;
+}