feat(curriculum): restore seed + solution to Chinese (#40683)

* feat(tools): add seed/solution restore script

* chore(curriculum): remove empty sections' markers

* chore(curriculum): add seed + solution to Chinese

* chore: remove old formatter

* fix: update getChallenges

parse translated challenges separately, without reference to the source

* chore(curriculum): add dashedName to English

* chore(curriculum): add dashedName to Chinese

* refactor: remove unused challenge property 'name'

* fix: relax dashedName requirement

* fix: stray tag

Remove stray `pre` tag from challenge file.

Signed-off-by: nhcarrigan <nhcarrigan@gmail.com>

Co-authored-by: nhcarrigan <nhcarrigan@gmail.com>

curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-35-circular-primes.md

@@ -3,6 +3,7 @@ id: 5900f38f1000cf542c50fea2
 title: 问题35：循环素数
 challengeType: 5
 videoUrl: ''
+dashedName: problem-35-circular-primes
 ---
 
 # --description--
@@ -47,5 +48,71 @@ assert(circularPrimes(750000) == 49);
 assert(circularPrimes(1000000) == 55);
 ```
 
+# --seed--
+
+## --seed-contents--
+
+```js
+function circularPrimes(n) {
+
+  return n;
+}
+
+circularPrimes(1000000);
+```
+
 # --solutions--
 
+```js
+function rotate(n) {
+  if (n.length == 1) return n;
+  return n.slice(1) + n[0];
+}
+
+function circularPrimes(n) {
+  // Nearest n < 10^k
+  const bound = 10 ** Math.ceil(Math.log10(n));
+  const primes = [0, 0, 2];
+  let count = 0;
+
+  // Making primes array
+  for (let i = 4; i <= bound; i += 2) {
+    primes.push(i - 1);
+    primes.push(0);
+  }
+
+  // Getting upperbound
+  const upperBound = Math.ceil(Math.sqrt(bound));
+
+  // Setting other non-prime numbers to 0
+  for (let i = 3; i < upperBound; i += 2) {
+    if (primes[i]) {
+      for (let j = i * i; j < bound; j += i) {
+        primes[j] = 0;
+      }
+    }
+  }
+
+  // Iterating through the array
+  for (let i = 2; i < n; i++) {
+    if (primes[i]) {
+      let curr = String(primes[i]);
+      let tmp = 1; // tmp variable to hold the no of rotations
+      for (let x = rotate(curr); x != curr; x = rotate(x)) {
+        if (x > n && primes[x]) {
+          continue;
+        }
+        else if (!primes[x]) {
+          // If the rotated value is 0 then it isn't a circular prime, break the loop
+          tmp = 0;
+          break;
+        }
+        tmp++;
+        primes[x] = 0;
+      }
+      count += tmp;
+    }
+  }
+  return count;
+}
+```