docs: add the other/miscellaneous form (#8163)

Co-authored-by: Christian Clauss <cclauss@me.com>
Co-authored-by: Dhruv Manilawala <dhruvmanila@gmail.com>

.github/ISSUE_TEMPLATE/other.yml

@@ -0,0 +1,19 @@
+name: Other
+description: Use this for any other issues. PLEASE do not create blank issues
+labels: ["awaiting triage"]
+body:
+  - type: textarea
+    id: issuedescription
+    attributes:
+      label: What would you like to share?
+      description: Provide a clear and concise explanation of your issue.
+    validations:
+      required: true
+
+  - type: textarea
+    id: extrainfo
+    attributes:
+      label: Additional information
+      description: Is there anything else we should know about this issue?
+    validations:
+      required: false

DIRECTORY.md

@@ -937,6 +937,8 @@
     * [Sol1](project_euler/problem_097/sol1.py)
   * Problem 099
     * [Sol1](project_euler/problem_099/sol1.py)
+  * Problem 100
+    * [Sol1](project_euler/problem_100/sol1.py)
   * Problem 101
     * [Sol1](project_euler/problem_101/sol1.py)
   * Problem 102

project_euler/problem_100/sol1.py

@@ -0,0 +1,48 @@
+"""
+Project Euler Problem 100: https://projecteuler.net/problem=100
+
+If a box contains twenty-one coloured discs, composed of fifteen blue discs and
+six red discs, and two discs were taken at random, it can be seen that
+the probability of taking two blue discs, P(BB) = (15/21) x (14/20) = 1/2.
+
+The next such arrangement, for which there is exactly 50% chance of taking two blue
+discs at random, is a box containing eighty-five blue discs and thirty-five red discs.
+
+By finding the first arrangement to contain over 10^12 = 1,000,000,000,000 discs
+in total, determine the number of blue discs that the box would contain.
+"""
+
+
+def solution(min_total: int = 10**12) -> int:
+    """
+    Returns the number of blue discs for the first arrangement to contain
+    over min_total discs in total
+
+    >>> solution(2)
+    3
+
+    >>> solution(4)
+    15
+
+    >>> solution(21)
+    85
+    """
+
+    prev_numerator = 1
+    prev_denominator = 0
+
+    numerator = 1
+    denominator = 1
+
+    while numerator <= 2 * min_total - 1:
+        prev_numerator += 2 * numerator
+        numerator += 2 * prev_numerator
+
+        prev_denominator += 2 * denominator
+        denominator += 2 * prev_denominator
+
+    return (denominator + 1) // 2
+
+
+if __name__ == "__main__":
+    print(f"{solution() = }")