Merge branch 'main' of http://www.wmarkj.com:3000/huanghai/python

TangDou/.vscode/settings.json

@@ -119,4 +119,5 @@
     "span": "cpp",
     "stop_token": "cpp"
   },
+  "C_Cpp.errorSquiggles": "disabled",
 }

TangDou/LanQiaoBei/LanQiao14STEMA202301/3.cpp

@@ -0,0 +1,23 @@
+#include <iostream>
+
+using namespace std;
+
+int func(int x) {
+    if (x <= 3)
+        return x * 2 - 1;
+    else if (x >= 6)
+        return func(x - 3) - 2;
+    else
+        return func(x + 1) + x;
+}
+
+int main() {
+    cout << func(14);
+    // f(14)=f(11)-2;
+    // f(11)=f(8)-2
+    // f(8)=f(5)-2
+    // f(5)=f(6)+5
+    // f(6)=f(3)-2
+    // f(3)=5
+    return 0;
+}

TangDou/LanQiaoBei/LanQiao14STEMA202301/5.cpp

@@ -0,0 +1,9 @@
+#include <bits/stdc++.h>
+using namespace std;
+// Hello world
+int main() {
+    char s[15];
+    cin >> s; // 原因：Hello world中存在空格，而cin会只读取前半部分Hello,所以长度是5
+    cout << strlen(s);
+    return 0;
+}

TangDou/LanQiaoBei/LanQiao14STEMA202301/BC2.cpp

@@ -0,0 +1,13 @@
+#include <iostream>
+using namespace std;
+int a[101];
+int main() {
+    int n;
+    cin >> n;
+    for (int i = 0; i < n; i++) cin >> a[i];
+
+    int s = a[0] + a[1] + a[2];
+    for (int i = 1; i < n - 2; i++) s = max(s, a[i] + a[i + 1] + a[i + 2]);
+    cout << s << endl;
+    return 0;
+}

TangDou/LanQiaoBei/LanQiao14STEMA202301/BC3.cpp

@@ -0,0 +1,36 @@
+#include <bits/stdc++.h>
+using namespace std;
+const int N = 120;
+int a[N], b[N];
+const int INF = 0x3f3f3f3f;
+/*
+3
+2 6 7
+*/
+int n, sum, avg;
+int main() {
+    cin >> n;
+    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]
+    }
+    avg = sum / n; // 平均数
+    int mi = INF;  // 最少次数，给初始值最大
+
+    for (int i = 0; i < n; i++) {                    // 按每个石子堆为起点
+        for (int j = 0; j < n; j++) b[j] = a[i + j]; // 从原始的破环成链数组中拷贝出来现在的临时数组
+
+        int cnt = 0;
+        for (int j = 0; j < n - 1; j++) {
+            cnt += abs(b[j] - avg); // 需要调整几个石子到下一个石子堆
+            b[j + 1] += b[j] - avg; // 下一个石子堆接着这些石子
+            b[j] = avg;             // j调整为平均值,其实这句没有发挥作用,可省略
+        }
+        // 以第i个石子为起点的讨论结束，判断是不是我们需要的最优解
+        mi = min(mi, cnt);
+    }
+
+    cout << mi << endl;
+    return 0;
+}

TangDou/LanQiaoBei/LanQiao14STEMA202301/题目.md

@@ -0,0 +1 @@
+https://ccgao.blog.csdn.net/article/details/134725679

TangDou/LanQiaoBei/LanQiao14STEMA20230212/10.cpp

@@ -0,0 +1,5 @@
+#include <bits/stdc++.h>
+using namespace std;
+int main() {
+    return 0;
+}

TangDou/LanQiaoBei/LanQiao14STEMA20230212/8.cpp

@@ -1,17 +1,23 @@
 #include <iostream>
 using namespace std;
+int n, cnt;
+
 int main() {
-    int n, cnt = 0;
     cin >> n;
-    cnt = n / 3;
     if (n == 2)
         cnt = -1;
     else if (n == 4)
         cnt = 4;
+    else if (n == 5)
+        cnt = 3;
+    else if (n == 7)
+        cnt = 3;
+    else if (n % 3 == 0)
+        cnt = n / 3;
     else if (n % 3 == 1)
-        cnt += 1;
+        cnt = (n - 7) / 3 + 3;
     else if (n % 3 == 2)
-        cnt += 2;
+        cnt = (n - 5) / 3 + 3;
 
     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;
+}

TangDou/LanQiaoBei/LanQiao14STEMA20230212/翻硬币.drawio

@@ -0,0 +1 @@
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