Add Tutorial For LOJ-1234 Harmonic Number (en) (#456)

* Added Tutorial LOJ-1234 (en)

* Added Tutorial

* Update en.md

Changed the solution approach

* Update en.md

Author: Talha-Taki002

1234/en.md

@@ -0,0 +1,55 @@
+# LOJ 1234 - Harmonic Number
+
+## Solution
+We won't be able to iterate from 1 to n for each test-case independently, since it will not fit the time limit. What we can do is iterate once from `n=1` to `1e8` (max) and compute `H(n)` for each of them using `H(n) = H(n-1) + 1/n`. Whenever we come across a `n` that is queried in the input, we will save the value for output. After the iteration has been completed, we will output all the queries together.
+
+### Complexity
+- Time Complexity: `O(MAX(N))`. 
+- Memory Complexity: `O(T)`.
+
+## C++ Code
+
+```cpp
+#include <bits/stdc++.h>
+
+using namespace std;
+
+
+int main() {
+    
+    // For fast I/O
+    ios_base::sync_with_stdio(false);
+    cin.tie(nullptr);
+
+    const int MAXN = 1e8;
+
+    int t;
+    cin >> t;
+
+    // 1-based indexing
+    vector <pair <int, int>> N(t+1);
+    for(int i = 1; i <= t; ++i) {
+        cin >> N[i].first;
+        // Saving corresponding index of the test case
+        N[i].second = i;
+    }
+    sort(N.begin(), N.end());
+
+    vector <double> ans(t+1);
+    double now = 0;
+    int idx = 1;
+    for(int i = 1; i <= MAXN; ++i) {
+        now += 1.0 / i;
+        // Could be duplicates, that's why using 'while' loop
+        while (idx < (int)N.size() && N[idx].first == i) {
+            ans[N[idx++].second] = now;
+        }
+    }
+
+    for(int ts = 1; ts <= t; ++ts) {
+        cout << fixed << setprecision(10) << "Case " << ts << ": " << ans[ts] << '\n';
+    }
+
+    return 0;
+}
+```