Merge pull request #437 from Krishna200608/feature/issue-406-rudolf-snowflakes

Solved Issue #406: Rudolf and Snowflakes (Hard Version)

Author: zonedoutdg

Problems/Binary-Search/Day-06/sol/Krishna200608/Solution2.cpp

@@ -0,0 +1,154 @@
+/*
+====================================================
+Problem: Rudolf and Snowflakes (Hard Version)
+Link: https://codeforces.com/contest/1846/problem/E2
+Author: Krishna Sikheriya (Krishna200608)
+====================================================
+
+-------------------------
+Short Problem Statement
+-------------------------
+You are given an integer n.
+Your task is to determine whether n can be written as a geometric series:
+
+    n = 1 + k + k² + ... + kᵖ
+
+where:
+- k ≥ 2 (the common ratio)
+- p ≥ 2 (so the series has at least 3 terms)
+
+In simpler terms, check if n can be formed by summing powers of some integer k,
+starting from 1, with at least three terms.
+
+-------------------------
+Approach (Length Fixing + Binary Search)
+-------------------------
+Key Observations:
+- The sum of a geometric series is:
+      n = (k^(p+1) - 1) / (k - 1)
+- Since n ≤ 10¹⁸, the maximum possible exponent is small.
+  For example, 2⁶⁰ already exceeds 10¹⁸.
+
+Strategy:
+1. Fix the number of terms in the series (terms = p + 1).
+   - Minimum terms = 3 (1 + k + k²)
+   - Maximum terms ≈ 60 (because 2⁶⁰ > 10¹⁸)
+2. For a fixed number of terms:
+   - The sum strictly increases as k increases.
+   - This allows binary search over possible values of k.
+3. For each candidate k, compute the geometric sum safely.
+   - Use __int128 to avoid overflow.
+   - Stop early if the sum exceeds n.
+4. If any combination produces exactly n, output "YES".
+   Otherwise, after all checks, output "NO".
+
+-------------------------
+Time & Space Complexity
+-------------------------
+Time Complexity:
+- Outer loop over possible terms: O(log n)
+- Binary search on k: O(log n)
+- Total per test case: O((log n)²)
+
+Space Complexity:
+- O(1), only constant extra space is used.
+
+-------------------------
+Example I/O (Optional)
+-------------------------
+Input:
+9
+1
+2
+3
+6
+13
+15
+255
+27
+4805
+
+Output:
+NO
+NO
+NO
+NO
+YES
+NO
+YES
+NO
+YES
+
+-------------------------
+Submission Link
+-------------------------
+https://codeforces.com/contest/1846/submission/356083531
+====================================================
+*/
+
+#include <bits/stdc++.h>
+using namespace std;
+
+// Safely computes: 1 + k + k^2 + ... + k^(terms - 1)
+// If the sum exceeds 'limit', returns a value greater than limit
+__int128 geometric_sum(__int128 k, int terms, __int128 limit) {
+    __int128 sum = 1;
+    __int128 current = 1;
+
+    for (int i = 1; i < terms; i++) {
+        // Prevent overflow while multiplying
+        if (current > limit / k) return limit + 1;
+        current *= k;
+
+        // Prevent overflow while adding
+        if (sum > limit - current) return limit + 1;
+        sum += current;
+    }
+    return sum;
+}
+
+void solve() {
+    long long n_input;
+    cin >> n_input;
+    __int128 n = n_input;
+
+    // Smallest valid snowflake: 1 + 2 + 4 = 7
+    if (n < 7) {
+        cout << "NO\n";
+        return;
+    }
+
+    // Try all possible lengths of the series (at least 3 terms)
+    for (int terms = 3; terms <= 62; terms++) {
+        long long low = 2;
+        long long high = 2000000000LL; // Safe upper bound for k
+
+        while (low <= high) {
+            long long mid = low + (high - low) / 2;
+            __int128 val = geometric_sum(mid, terms, n);
+
+            if (val == n) {
+                cout << "YES\n";
+                return;
+            } else if (val < n) {
+                low = mid + 1;
+            } else {
+                high = mid - 1;
+            }
+        }
+    }
+
+    cout << "NO\n";
+}
+
+int main() {
+    ios::sync_with_stdio(false);
+    cin.tie(nullptr);
+
+    int t;
+    cin >> t;
+    while (t--) {
+        solve();
+    }
+    return 0;
+}