Converted everything to orgmode

converted everything to orgmode and added solution to the README files

Informatik_I/Exercise_2/task_4/README.md

@@ -1,14 +0,0 @@
-## Task
-
-Write a program that inputs a non-negative integer `n` (but store it as `int`) and outputs the binary digits of `n` in the _correct_ order (i.e., starting with the most significant bit). Do not output the leading zeros or the sign.
-
-**Hint:** In order to find the largest integer $k$ such that $2^k \leq x$, you can utilize that $k$ is the smallest integer such that $2^k > \frac{x}{2}$. This observation is particularly useful to avoid an overflow for the expression $2^k$ when searching for the most significant bit to represent $x$.
-
-**Restrictions:**
-
-- Libraries: only the iostream standard library header is allowed; using arrays, string or cmath is not permitted.
-- Operators: you may not use bitshift operators to manipulate the numbers.
-
----
-
-**Warning:** The autograder does not catch if you do anything that's disallowed in the "Restrictions" section above, but you will receive 0 points when your TA corrects and catches this.

Informatik_I/Exercise_2/task_4/README.org

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+#+TITLE: Task 4: From decimal to binary representation
+#+AUTHOR: JirR02
+
+* Task
+
+Write a program that inputs a non-negative integer `n` (but store it as `int`) and outputs the binary digits of `n` in the /correct/ order (i.e., starting with the most significant bit). Do not output the leading zeros or the sign.
+
+*Hint:* In order to find the largest integer \(k\) such that \(2^k \leq x\), you can utilize that \(k\) is the smallest integer such that \(2^k > \frac{x}{2}\). This observation is particularly useful to avoid an overflow for the expression \(2^k\) when searching for the most significant bit to represent \(x\).
+
+*Restrictions:*
+
+- Libraries: only the iostream standard library header is allowed; using arrays, string or cmath is not permitted.
+- Operators: you may not use bitshift operators to manipulate the numbers.
+
+-----
+
+*Warning:* The autograder does not catch if you do anything that's disallowed in the "Restrictions" section above, but you will receive 0 points when your TA corrects and catches this.
+
+* Solution
+
+#+begin_src cpp
+#include <iostream>
+
+int main() {
+  int i = 0; // Input integer
+  int d = 0; // Input Number
+  int a = 0; // Part of binary number
+  int b = 0; // Number to divide
+  int count = 0;
+
+  std::cin >> i;
+  d = i;
+  b = i;
+
+  if (i < 0) {
+    return 0;
+  } else if (i == 0) {
+    std::cout << 0;
+  } else {
+    while (d != 0) {
+      d = d / 2;
+      ++count;
+    }
+    for (; count > 0; --count) {
+      for (int j = count; j > 1; --j) {
+        b = b / 2;
+      }
+      a = b % 2;
+      std::cout << a;
+      b = i;
+    }
+  }
+}
+#+end_src

Informatik_I/Exercise_2/task_4/main.cpp

@@ -1,32 +0,0 @@
-#include <iostream>
-
-int main() {
-  int i = 0; // Input integer
-  int d = 0; // Input Number
-  int a = 0; // Part of binary number
-  int b = 0; // Number to divide
-  int count = 0;
-
-  std::cin >> i;
-  d = i;
-  b = i;
-
-  if (i < 0) {
-    return 0;
-  } else if (i == 0) {
-    std::cout << 0;
-  } else {
-    while (d != 0) {
-      d = d / 2;
-      ++count;
-    }
-    for (; count > 0; --count) {
-      for (int j = count; j > 1; --j) {
-        b = b / 2;
-      }
-      a = b % 2;
-      std::cout << a;
-      b = i;
-    }
-  }
-}