'commit'

TangDou/LanQiaoBei/LanQiao14STEMA20230212/题解.md

@@ -1,3 +0,0 @@
-https://blog.csdn.net/qq_36230375/article/details/134725732
-
-https://tiku.scratchor.com/paper/view/gjfvsjrgtpm30v3f

TangDou/LanQiaoBei/ZhongGaoJi/LanQiao14STEMA202301/BC3.cpp

@@ -4,8 +4,8 @@ const int N = 120;
 int a[N], b[N];
 const int INF = 0x3f3f3f3f;
 /*
-3
+5
-2 6 7
+5 7 5 5 3
 */
 int n, sum, avg;
 int main() {
@@ -13,7 +13,7 @@ int main() {
     for (int i = 0; i < n; i++) {
         cin >> a[i];
         sum += a[i];
-        a[i + n] = a[i]; // 破环成链, 0 1 2-->a[3]=a[0],a[4]=a[1],a[5]=a[2]
+        a[i + n] = a[i]; // 破环成链
     }
     avg = sum / n; // 平均数
     int mi = INF;  // 最少次数，给初始值最大

TangDou/LanQiaoBei/ZhongGaoJi/LanQiao14STEMA202301/BC4.cpp

@@ -0,0 +1,33 @@
+#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++) {     // 枚举每个小正方形的边长
+        for (int i = 0; i <= n - e; i++) {     // 起点i
+            for (int j = 0; j <= m - e; j++) { // 起点j
+                bool f = true;
+                for (int x = 0; x < e; x++) { // 遍历正方形中所有的其它位置，判断是不是与左上角的一致
+                    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/LanQiao14STEMA202301/BC5.cpp

@@ -0,0 +1,69 @@
+#include <bits/stdc++.h>
+using namespace std;
+const int INF = 0x3f3f3f3f;
+const int N = 110;
+double a[N];
+int n;
+int f1[N], f2[N];
+int main() {
+    cin >> n;
+    for (int i = 1; i <= n; i++) cin >> a[i];
+    // 以每棵树为保留的，并且是最高的那棵树，分别求出最长上升序列长度x 和 最长下降序列长度y， x+y-1就是最终保留的整体序列长度，
+    // 去掉的数量就是 n-(x+y-1)的值。然后求min()
+    // 需要注意的是默认值设置为-1，比如 5 4 3 2 1 ,我们取哪棵为最高点都行不通，不存在左侧上侧到峰值的情况，右侧即使符合也不行的。
+
+    // 求最长上升
+    for (int i = 1; i <= n; i++) {
+        f1[i] = 1;
+        for (int j = 1; j < i; j++)
+            if (a[i] > a[j]) f1[i] = max(f1[i], f1[j] + 1);
+    }
+    // 求最长下降
+    for (int i = n; i >= 1; i--) {
+        f2[i] = 1;
+        for (int j = n; j > i; j--)
+            if (a[i] > a[j]) f2[i] = max(f2[i], f2[j] + 1);
+    }
+
+    // // 输出最长上升
+    // for (int i = 1; i <= n; i++) cout << f1[i] << " ";
+    // cout << endl;
+
+    // // 输出最长下降
+    // for (int i = 1; i <= n; i++) cout << f2[i] << " ";
+    // cout << endl;
+
+    bool flag = true;
+    for (int i = 1; i <= n; i++)
+        if (f1[i] != 1) {
+            flag = false;
+            break;
+        }
+
+    if (flag) {
+        cout << -1 << endl;
+        exit(0);
+    }
+
+    flag = true;
+    for (int i = 1; i <= n; i++)
+        if (f2[i] != 1) {
+            flag = false;
+            break;
+        }
+
+    if (flag) {
+        cout << -1 << endl;
+        exit(0);
+    }
+
+    int mi = -1;
+    for (int i = 1; i <= n; i++) {
+        int tmp = f1[i] + f2[i];
+        mi = max(mi, tmp);
+    }
+
+    cout << n - mi + 1 << endl;
+
+    return 0;
+}

TangDou/LanQiaoBei/ZhongGaoJi/LanQiao14STEMA202301/BC5.in

@@ -0,0 +1,2 @@
+10
+1.0 2.3 1.2 1.7 1.1 2.0 1.8 1.8 1.2 1.9

TangDou/LanQiaoBei/ZhongGaoJi/LanQiao14STEMA202302/9.cpp

@@ -21,7 +21,7 @@ int main() {
         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;
+    int res = 0;
     for (int x1 = 1; x1 <= n; x1++)
         for (int y1 = 1; y1 <= m; y1++)
             for (int x2 = x1; x2 <= n; x2++)

TangDou/LanQiaoBei/ZhongGaoJi/LanQiao14STEMA202302/DLL.CPP

@@ -0,0 +1,11 @@
+#include <bits/stdc++.h>
+using namespace std;
+int main() {
+    int n;
+    cin >> n;
+    n %= 100;
+    n = n - n % 10;
+    n /= 10;
+    cout << n;
+    return 0;
+}

TangDou/LanQiaoBei/ZhongGaoJi/LanQiao14STEMA202302/题解.md

@@ -0,0 +1 @@
+https://ccgao.blog.csdn.net/article/details/134725732

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/题解.md

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