int main() ios::sync_with_stdio(false); cin.tie(nullptr); int T; if (!(cin >> T)) return 0; while (T--) int N; cin >> N; vector<int> love(N + 1); // 1‑based for (int i = 1; i <= N; ++i) cin >> love[i];
love[1 … N] // 1‑based indexing where love[i] = j means person i loves person j .
A is an unordered pair i , j ( i ≠ j ) such that
(A classic “mutual‑love” counting problem – often seen on SPOJ, LightOJ, and other online judges) 1️⃣ Problem statement You are given a group of N people, numbered from 1 to N . Each person loves exactly one other person (possibly himself). The love‑relationships are described by an array
From Lemma 1 every increment corresponds to a genuine mutual‑love pair. From Lemma 2 every genuine pair contributes exactly one increment. From Lemma 3 no non‑mutual pair contributes any increment. Therefore the total number of increments equals precisely the number of mutual‑love pairs. ∎ 5️⃣ Complexity analysis Time – The loop visits each of the N people once, performing O(1) work per iteration: O(N) per test case.
When the loop later reaches i = b , the first condition fails ( b < a is false), so the pair is counted again. ∎ Lemma 3 If a pair i, j is not a mutual‑love pair, the algorithm never increments mutualPairs for it.
import sys
love[i] = j and love[j] = i . Your task is to count how many mutual‑love pairs exist in the given group.
2 4 2 1 4 3 5 2 3 1 5 4
Memory – The array love[1…N] is stored: .
Both limits satisfy the given constraints ( ∑ N ≤ 10⁶ ). Below are clean, production‑ready solutions in C++ (17) and Python 3 . Both follow the algorithm described above and use fast I/O to handle the maximum input size. C++ (GNU‑C++17) #include <bits/stdc++.h> using namespace std;