'commit'

TangDou/LanQiaoBei/ZhongGaoJi/LanQiao14STEMA202303/1.cpp

@@ -0,0 +1,6 @@
+#include <bits/stdc++.h>
+using namespace std;
+int main() {
+    cout << '9' * 3 << endl;
+    return 0;
+}

TangDou/LanQiaoBei/ZhongGaoJi/LanQiao14STEMA202303/4.cpp

@@ -0,0 +1,7 @@
+#include <bits/stdc++.h>
+using namespace std;
+int main() {
+    int b = 0x212;
+    cout << b;
+    return 0;
+}

TangDou/LanQiaoBei/ZhongGaoJi/LanQiao14STEMA202303/BC1.cpp

@@ -0,0 +1,8 @@
+#include <bits/stdc++.h>
+using namespace std;
+int main() {
+    int n;
+    cin >> n;
+    cout << n - n / 200 * 25;
+    return 0;
+}

TangDou/LanQiaoBei/ZhongGaoJi/LanQiao14STEMA202303/BC2.cpp

@@ -0,0 +1,15 @@
+#include <bits/stdc++.h>
+using namespace std;
+const int N = 10010;
+int n, mx, pre, nxt;
+
+int main() {
+    cin >> n >> pre;
+    for (int i = 2; i <= n; i++) {
+        cin >> nxt;
+        if (abs(nxt - pre) > mx) mx = abs(nxt - pre);
+        pre = nxt;
+    }
+    cout << mx;
+    return 0;
+}

TangDou/LanQiaoBei/ZhongGaoJi/LanQiao14STEMA202303/BC3.cpp

@@ -0,0 +1,37 @@
+#include <bits/stdc++.h>
+using namespace std;
+
+int check(int x) {
+    // 转9进制
+    string res;
+    while (x) {
+        int a = x % 9;
+        res += to_string(a);
+        x /= 9;
+    }
+    // 每个数位都是奇数
+    for (int i = 0; i < res.size(); i++)
+        if ((res[i] - '0') % 2 == 0) return 0;
+
+    // 判断转完的9进制是不是回文数
+    // 方法1
+    // string str = res;
+    // reverse(str.begin(), str.end());
+    // return str == res;
+
+    // 方法2
+    for (int i = 0; i < (res.size() + 1) / 2; i++)
+        if (res[i] != res[res.size() - 1 - i]) return 0;
+
+    return 1;
+}
+int main() {
+    int n, m;
+    cin >> n >> m;
+
+    int res = 0;
+    for (int i = n; i <= m; i++) res += check(i);
+
+    cout << res << endl;
+    return 0;
+}

TangDou/LanQiaoBei/ZhongGaoJi/LanQiao14STEMA202303/BC4.cpp

@@ -0,0 +1,42 @@
+#include <bits/stdc++.h>
+using namespace std;
+const int N = 110;
+int n, m, ans, st[N];
+int a[N][N];
+
+bool check(int u) {
+    for (int i = 1; i < u; i++)
+        if (st[i] && a[u][i]) return false;
+    return true;
+}
+
+void dfs(int u, int sum) {   // 第u号钻石，之前一共选了sum个钻石
+    if (u == n + 1) {        // 如果所有钻石都找完了
+        ans = max(ans, sum); // 取最大值
+        return;
+    }
+    // 如果剩下的钻石都选也没有已有的最好情况多，返回,剪枝
+    // if (ans > sum + (n - u + 1)) return;
+
+    if (check(u)) {          // 当前钻石与已选钻石没有冲突
+        st[u] = 1;           // 选之
+        dfs(u + 1, sum + 1); // 继续下一个钻石
+        st[u] = 0;           // 不选
+    }
+
+    dfs(u + 1, sum); // 继续下一个钻石
+}
+
+int main() {
+    cin >> n >> m;
+    while (m--) {
+        int x, y;
+        cin >> x >> y;
+        a[x][y] = a[y][x] = 1; // x<->y互相冲突
+    }
+
+    dfs(1, 0); // 从第一个钻石开始选，目前已选0个
+    cout << ans << endl;
+
+    return 0;
+}

TangDou/LanQiaoBei/ZhongGaoJi/LanQiao14STEMA202303/ggsddu.cpp

@@ -0,0 +1,31 @@
+#include <bits/stdc++.h>
+using namespace std;
+const int N = 31;
+char a[N][N];
+int n, m, cnt;
+
+int main() {
+    cin >> n >> m;
+    for (int i = 0; i < n; i++)
+    for (int j = 0; j < m; j++)
+        cin >> a[i][j];
+    for (int e = 2; e <= min(m, n);e++){
+        for (int i = 0; i <= n - e;i++){
+            for (int j = 0; j <= m - e;j++){
+                bool f = true;
+                for (int x = 0; x < e;x++){
+                    for (int y = 0; y < e;y++){
+                        if(a[i][j]!=a[i+x][j+y]){
+                            f = false;
+                            break;
+                        }
+                    }
+                    if (!f) break;
+                }
+                if (f) cnt++;
+            }
+        }
+    }
+    cout << cnt << endl;
+    return 0;
+}

TangDou/LanQiaoBei/ZhongGaoJi/LanQiao14STEMA202303/题解.md

@@ -0,0 +1,3 @@
+https://ccgao.blog.csdn.net/article/details/134725794
+
+https://tiku.scratchor.com/paper/view/z8k4y0xsf0ue8xlh