Added Tutorial LOJ-1234

1234/en.md

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+# LOJ 1234 - Harmonic Number
+
+## Solution
+From the definition of H<sub>n</sub> we perceive: **H<sub>n+1</sub> = H<sub>n</sub> + (1/ n)**
+
+So we need not to calculate **H<sub>n</sub>** using the straighforward formula for every single test case. We go from non-decreasing order of test cases whilst keeping a variable for harmonic sum so far. This way not only we save time but memory as well, because there's no need for storing all the harmonic values beforehand.
+
+## Complexity
+- Time Complexity: O($T$). 
+- Memory Complexity: O($T$).
+
+## Code
+
+### C++
+
+```cpp
+#include <bits/stdc++.h>
+
+using namespace std;
+
+
+typedef pair <int, int> pii;
+
+
+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 <pii> 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. / i;
+        while (idx < (int)N.size() && N[idx].first == i) { // Could be duplicates, that's why using 'while' loop
+            ans[N[idx++].second] = now;
+        }
+    }
+
+    for(int ts = 1; ts <= t; ++ts) {
+        cout << fixed << setprecision(10) << "Case " << ts << ": " << ans[ts] << '\n';
+    }
+
+    return 0;
+}
+```