Merge pull request #424 from Aaryan-Degama/d6q2

D6q2

Author: Dheeraj-06

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

@@ -0,0 +1,83 @@
+// link : https://codeforces.com/contest/1846/submission/356025360
+
+#include <iostream>
+#include <vector>
+#include <cmath>
+#include <cstdint>
+#include <algorithm>
+
+using namespace std;
+
+bool check(long long k, int m, long long n) {
+    __int128 sum = 1;
+    __int128 cur = 1;
+
+    for (int i = 1; i <= m; i++) {
+        cur *= k;
+        sum += cur;
+        if (sum > n) return false;
+    }
+    return sum == n;
+}
+
+int main() {
+    ios::sync_with_stdio(false);
+    cin.tie(nullptr);
+
+    int t;
+    cin >> t;
+
+    while (t--) {
+        long long n;
+        cin >> n;
+
+        bool possible = false;
+
+        for (int m = 2; m <= 60 && !possible; m++) {
+            long long low = 2, high = 1e9;
+
+            while (low <= high) {
+                long long mid = (low + high) / 2;
+
+                if (check(mid, m, n)) {
+                    possible = true;
+                    break;
+                }
+
+                __int128 sum = 1, cur = 1;
+                for (int i = 1; i <= m; i++) {
+                    cur *= mid;
+                    sum += cur;
+                    if (sum > n) break;
+                }
+
+                if (sum < n)
+                    low = mid + 1;
+                else
+                    high = mid - 1;
+            }
+        }
+
+        cout << (possible ? "YES\n" : "NO\n");
+    }
+
+    return 0;
+}
+
+
+/*
+Explaination:
+Try to represent n as a geometric series:
+    n = 1 + k + k^2 + ... + k^m , where k >= 2 and m >= 2.
+Since powers grow fast, we only need to check m up to 60.
+For each m, try different values of k using binary search.
+For every (k, m), compute the sum iteratively and stop early if it exceeds n.
+If any sum equals n, output "YES"; otherwise, output "NO".
+
+Time Complexity:
+O(n)
+
+Space Complexity:
+O(1)
+*/
+