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/rosetta-code/gaussian-elimination.md

@@ -3,6 +3,7 @@ id: 5a23c84252665b21eecc7e77
 title: 高斯消除
 challengeType: 5
 videoUrl: ''
+dashedName: gaussian-elimination
 ---
 
 # --description--
@@ -109,5 +110,117 @@ assert.deepEqual(
 );
 ```
 
+# --seed--
+
+## --seed-contents--
+
+```js
+function gaussianElimination(A,b) {
+
+}
+```
+
 # --solutions--
 
+```js
+function gaussianElimination(A, b) {
+  // Lower Upper Decomposition
+  function ludcmp(A) {
+    // A is a matrix that we want to decompose into Lower and Upper matrices.
+    var d = true
+    var n = A.length
+    var idx = new Array(n) // Output vector with row permutations from partial pivoting
+    var vv = new Array(n) // Scaling information
+
+    for (var i=0; i<n; i++) {
+        var max = 0
+        for (var j=0; j<n; j++) {
+            var temp = Math.abs(A[i][j])
+            if (temp > max) max = temp
+        }
+        if (max == 0) return // Singular Matrix!
+        vv[i] = 1 / max // Scaling
+    }
+
+        var Acpy = new Array(n)
+        for (var i=0; i<n; i++) {
+            var Ai = A[i]
+            let Acpyi = new Array(Ai.length)
+            for (j=0; j<Ai.length; j+=1) Acpyi[j] = Ai[j]
+            Acpy[i] = Acpyi
+        }
+        A = Acpy
+
+    var tiny = 1e-20 // in case pivot element is zero
+    for (var i=0; ; i++) {
+        for (var j=0; j<i; j++) {
+            var sum = A[j][i]
+            for (var k=0; k<j; k++) sum -= A[j][k] * A[k][i];
+            A[j][i] = sum
+        }
+        var jmax = 0
+        var max = 0;
+        for (var j=i; j<n; j++) {
+            var sum = A[j][i]
+            for (var k=0; k<i; k++) sum -= A[j][k] * A[k][i];
+            A[j][i] = sum
+            var temp = vv[j] * Math.abs(sum)
+            if (temp >= max) {
+                max = temp
+                jmax = j
+            }
+        }
+        if (i <= jmax) {
+            for (var j=0; j<n; j++) {
+                var temp = A[jmax][j]
+                A[jmax][j] = A[i][j]
+                A[i][j] = temp
+            }
+            d = !d;
+            vv[jmax] = vv[i]
+        }
+        idx[i] = jmax;
+        if (i == n-1) break;
+        var temp = A[i][i]
+        if (temp == 0) A[i][i] = temp = tiny
+        temp = 1 / temp
+        for (var j=i+1; j<n; j++) A[j][i] *= temp
+    }
+    return {A:A, idx:idx, d:d}
+  }
+
+  // Lower Upper Back Substitution
+  function lubksb(lu, b) {
+    // solves the set of n linear equations A*x = b.
+    // lu is the object containing A, idx and d as determined by the routine ludcmp.
+    var A = lu.A
+    var idx = lu.idx
+    var n = idx.length
+
+        var bcpy = new Array(n)
+        for (var i=0; i<b.length; i+=1) bcpy[i] = b[i]
+        b = bcpy
+
+    for (var ii=-1, i=0; i<n; i++) {
+        var ix = idx[i]
+        var sum = b[ix]
+        b[ix] = b[i]
+        if (ii > -1)
+            for (var j=ii; j<i; j++) sum -= A[i][j] * b[j]
+        else if (sum)
+            ii = i
+        b[i] = sum
+    }
+    for (var i=n-1; i>=0; i--) {
+        var sum = b[i]
+        for (var j=i+1; j<n; j++) sum -= A[i][j] * b[j]
+        b[i] = sum / A[i][i]
+    }
+    return b // solution vector x
+  }
+
+    var lu = ludcmp(A)
+    if (lu === undefined) return // Singular Matrix!
+    return lubksb(lu, b)
+}
+```