USACO 2022 January Contest Silver Division - Searching for Soulmates

Problem link: here

Solution Author: Stefan Dascalescu

Problem Solution

We notice that it takes no more than \(\log x\) operations to bring \(x\) to \(1\). We create two vectors (one for A and one for B) that count how many operations it takes to bring \(x\) to state \(x'\) and take all the combinations \((B'-A'+i+j)\), and the minimum of these is the answer.

Source code

The source code in C++ can be seen below.

#include <climits>
#include <fstream>
#include <iostream>
#include <vector>
using namespace std;
int t;
void solve() {
    long long a, b;
    cin >> a >> b;
    vector<long long> div1 = {a}, div2 = {b};
    while (a > 1) {
        if (a % 2) {
            a++;
        } else
            a /= 2;
        div1.push_back(a);
    }
    while (b > 1) {
        if (b % 2) {
            b--;
        } 
        else
            b /= 2;
        div2.push_back(b);
    }
    long long ret = LLONG_MAX;
    for (int i = 0; i < div1.size(); i++) {
        for (int j = 0; j < div2.size(); j++) {
            if (div1[i] <= div2[j]) {
                ret = min(ret, div2[j] - div1[i] + i + j);
            }
        }
    }
    cout << ret << "\n";
    return;
}
int main() {
    cin >> t;
    while (t--) {
        solve();
    }
    return 0;
}