add-day1-sol

Author: codeboy-pro

Problems/Mathematics/Day-01/sol/PradipMaity/Solution1.cpp

@@ -0,0 +1,61 @@
+/*
+Problem:
+People are standing in a circle numbered 1 to n (n is even).
+Person a is opposite person b.
+We must find who is opposite person c.
+If impossible, print -1.
+
+Key Observation:
+If a is opposite b → distance between them is exactly n/2.
+So:
+    n = |a - b| * 2
+
+Once we know n, we know half circle size = n / 2
+
+Opposite of any person x is:
+    x + n/2      (if <= n)
+otherwise:
+    x - n/2
+
+Invalid Cases:
+1.max(a, b, c) > n  → means c doesn't exist in circle
+2.a == b or diff == 0   → opposite cannot be same
+Output -1 in such cases.
+*/
+
+#include <bits/stdc++.h>
+using namespace std;
+
+int main() {
+    int t;
+    cin >> t;
+
+    while(t--){
+        long long a, b, c;
+        cin >> a >> b >> c;
+
+        // Distance between a and b
+        long long diff = abs(a - b);
+
+        // Total number of people in circle
+        long long n = diff * 2;
+
+        // Invalid cases
+        if (diff == 0 || a > n || b > n || c > n) {
+            cout << -1 << endl;
+            continue;
+        }
+
+        // Halfway around the circle
+        long long half = n / 2;
+
+        // Opposite of c
+        long long ans = c + half;
+
+        // If goes beyond circle wrap around
+        if (ans > n) ans -= n;
+
+        cout << ans << endl;
+    }
+    return 0;
+}

Problems/Mathematics/Day-01/sol/PradipMaity/Solution2.cpp

@@ -0,0 +1,76 @@
+/*
+We count pairs (i, j) such that ai * aj is a perfect k-th power.
+
+Idea:
+For each number, compute its prime factorization.
+Reduce each exponent modulo k because only residue matters.
+This forms a "signature".
+
+Example (k = 3)
+24 = 2^3 * 3^1  → exponents mod 3 → 2^0 * 3^1
+
+To pair with another number to form perfect k-th power,
+for each prime exponent 'e', we need (k - e) % k.
+
+So for each number, we compute:
+  signature        = reduced exponents
+  neededSignature  = complementary exponents
+
+We store counts of signatures in a map.
+For each number, answer += count[neededSignature].
+
+Time Complexity:
+O(n * sqrt(max(ai)))
+
+Space Complexity:
+O(n)
+*/
+
+#include <bits/stdc++.h>
+using namespace std;
+
+int main() {
+    ios::sync_with_stdio(false);
+    cin.tie(nullptr);
+
+    int n, k;
+    cin >> n >> k;
+
+    vector<int> a(n);
+    for (int &x : a) cin >> x;
+
+    map<vector<pair<int,int>>, long long> freq;
+    long long ans = 0;
+
+    for (int x : a) {
+        int num = x;
+        map<int,int> cnt;
+
+        // Prime factorize
+        for (int p = 2; p * p <= num; p++) {
+            while (num % p == 0) {
+                cnt[p]++;
+                num /= p;
+            }
+        }
+        if (num > 1) cnt[num]++;
+
+        vector<pair<int,int>> sig;   // reduced signature
+        vector<pair<int,int>> need;  // complementary signature
+
+        for (auto &it : cnt) {
+            int prime = it.first;
+            int e = it.second % k;
+            if (e != 0) {
+                sig.push_back({prime, e});
+                need.push_back({prime, (k - e) % k});
+            }
+        }
+
+        ans += freq[need];
+        freq[sig]++;
+    }
+
+    cout << ans << "\n";
+    return 0;
+}