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var_e vs var_f

Problem: two-sum

A→B9%B→A9%Shared fingerprints5
24tokens in the longest matched region
var_e
#include <bits/stdc++.h>
using namespace std;

// two-sum, hash map approach, O(n) expected

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);

    int total;          // element count
    long long goal;     // required sum
    cin >> total >> goal;

    unordered_map<long long, int> prior;    // value -> earliest index
    prior.reserve(2 * total);

    int pos = 1;
    while (pos <= total) {
        long long cur;
        cin >> cur;                          // next value
        long long complement = goal - cur;
        auto hit = prior.find(complement);
        if (prior.end() != hit) {
            // found the partner we stored earlier
            cout << hit->second << ' ' << pos << '\n';
            return 0;
        }
        prior[cur] = pos;                    // stash for later
        pos = pos + 1;
    }
    return 0;
}
var_f
#include <bits/stdc++.h>
using namespace std;

static int runs = 0;

void banner() {
    // deliberately original code around the lifted part
    runs++;
}

// ---- lifted from the original solution (this is the copied part) ----
int solve(int n, long long t) {
    unordered_map<long long, int> seen;
    seen.reserve(n * 2);
    for (int idx = 1; idx <= n; idx++) {
        long long x;
        cin >> x;
        auto it = seen.find(t - x);
        if (it != seen.end()) {
            cout << it->second << ' ' << idx << '\n';
            return 0;
        }
        seen[x] = idx;
    }
    return 1;
}
// ---------------------------------------------------------------------

int main() {
    banner();
    int n; long long t;
    if (!(cin >> n >> t)) return 1;
    int status = solve(n, t);
    banner();
    return status == 0 ? 0 : 2;
}