1152C - Neko does Maths - CodeForces Solution

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>

#pragma GCC optimize("O3")
#pragma GCC target("avx,avx2,fma")
#pragma GCC optimization ("unroll-loops")
#pragma GCC target("avx,avx2,sse,sse2,sse3,sse4,popcnt")

using namespace std;
using namespace __gnu_pbds;
#define int long long
#define float long double
#define elif else if
#define endl "\n"
#define mod 1000000007
#define pi acos(-1)
#define eps 0.000000001
#define inf 1000000000000000000
#define FIXED(a) cout << fixed << setprecision(a)
#define all(x) x.begin(), x.end()
#define rall(x) x.rbegin(), x.rend()
#define time_init auto start = std::chrono::high_resolution_clock::now()
#define time_report auto end = std::chrono::high_resolution_clock::now(); std::cout << std::chrono::duration_cast<std::chrono::milliseconds>(end - start).count() << " ms" << endl
#define debug(x) { cerr << #x << " = " << x << endl; }
#define len(x) (int)x.size()
#define sqr(x) ((x) * (x))
#define cube(x) ((x) * (x) * (x))
#define bit(x, i) (((x) >> (i)) & 1)
#define set_bit(x, i) ((x) | (1LL << (i)))
#define clear_bit(x, i) ((x) & (~(1LL << (i))))
#define toggle_bit(x, i) ((x) ^ (1LL << (i)))
#define low_bit(x) ((x) & (-(x)))
#define count_bit(x) __builtin_popcountll(x)
#define srt(x) sort(all(x))
#define rsrt(x) sort(rall(x))

template<typename T> using min_heap = priority_queue<T, vector<T>, greater<T>>;
template<typename T> using max_heap = priority_queue<T, vector<T>, less<T>>;
template<typename T> using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
template<typename T> using ordered_multiset = tree<T, null_type, less_equal<T>, rb_tree_tag, tree_order_statistics_node_update>;
template<typename T> using matrix = vector<vector<T>>;
template<typename T> using graph = vector<vector<T>>;


signed main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);
    int a, b;
    cin >> a >> b;
    if (a > b)
        swap(a, b);
    int diff = b - a;
    int ans = a * b / (__gcd(a, b));
    int kans = 0;
    for (int i = 1; i * i <= diff; ++i) {
        if (diff % i == 0) {
            int x = i;
            int newb = 0;
            if (b % x == 0)
                newb = b;
            else
                newb = b + x - b % x;
            int k = newb - b;
            int val = (a + k) * (b + k) / (__gcd(a + k, b + k));
            if (val < ans) {
                ans = val;
                kans = k;
            }

            x = diff / i;
            if (x == i)
                continue;
            if (b % x == 0)
                newb = b;
            else
                newb = b + x - b % x;
            k = newb - b;
            val = (a + k) * (b + k) / (__gcd(a + k, b + k));
            if (val < ans) {
                ans = val;
                kans = k;
            }
        }
    }
    cout << kans << endl;
}