Basic Nim Game

1. Calculate xor_sum of all piles.
2. If xor_sum == 0:
   1. Print that the second player wins.
3. Otherwise:
   1. Find a pile i such that (a[i] xor xor_sum) < a[i].
   2. Set new_size = a[i] xor xor_sum.
   3. Print that the first player wins and the move i, new_size.

#include <bits/stdc++.h>

#define fi first
#define se second

using namespace std;

void solve() {
    int n;
    cin >> n;

    vector<long long> a(n);
    long long xr = 0;
    for (long long &x : a) {
        cin >> x;
        xr ^= x;
    }

    if (xr == 0) {
        cout << "SECOND\n";
        return;
    }

    cout << "FIRST\n";
    for (int i = 0; i < n; i++) {
        long long target = a[i] ^ xr;
        if (target < a[i]) {
            cout << i + 1 << ' ' << target << '\n';
            return;
        }
    }
}

int32_t main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);

    solve();

    return 0;
}

Introduction

Problem or motivation

Core concept

Nim Game Variants

Main algorithm

Mental model and detailed walkthrough

Implementation

Common mistakes

Complexity