# 地址

https://codeforces.com/contest/1152/problem/C

# 原文地址

https://www.lucien.ink/archives/423/

## 题解

$$lcm(a + k, b + k)$$$$= \frac{(a + k) \cdot (b + k)}{gcd(a + k, b+ k)}$$$$= \frac{a \cdot b + (a + b) \cdot k + k ^ 2}{gcd(b + k, a - b)}$$

## 代码

https://pasteme.cn/6827

#include <bits/stdc++.h>
typedef long long ll;
ll gcd(ll p, ll q) { return q ? gcd(q, p % q) : p; }
ll a, b, ans_k{}, ans = 0x3f3f3f3f3f3f3f3f;
ll calc(ll a, ll b) { // find minimal k, st. b * k >= a, then return b * k
if (a <= b) return b;
if (a % b == 0) return a;
return (a + b - a % b);
}
std::vector<ll> fac;
int main() {
scanf("%lld%lld", &a, &b);
if (a == b) return 0 * puts("0");
if (a < b) std::swap(a, b);
ans = a * b / gcd(a, b);
ll upper = calc(b, a - b);
for (ll i = 1; i * i <= upper; i++) {
if (upper % i == 0) {
fac.push_back(i);
fac.push_back(upper / i);
}
}
for (auto each : fac) {
ll k = calc(b, each) - b, tmp = (a * b + (a + b) * k + k * k) / each;
if (tmp < ans) {
ans = tmp;
ans_k = k;
}
}
printf("%lld\n", ans_k);
return 0;
}