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

题意：给两个数a和b，找到一个最小的k，使得a+k和b+k的lcm最小

题解：假设a>b,gcd(a,b)=gcd(b,a-b)，所以我们要求的gcd(a+k,b+k)可以转化成gcd(b+k,a-b)，令d=gcd(b+k,a-b)，则d是a-b的因数，枚举这个因数，枚举过程中不断更新最小的lcm对应的k，若lcm相等，就直接更新最小k

代码：

 1 //#include<bits/stdc++.h>
 2 #include<iostream>
 3 #include<vector>
 4 #include<stack>
 5 #include<string>
 6 #include<cstdio>
 7 #include<algorithm>
 8 #include<queue>
 9 #include<map>
10 #include<set>
11 #include<cmath>
12 #include<iomanip>
13 #define inf 0x3f3f3f3f
14 using namespace std;
15 typedef long long ll;
16 const int M = int(1e7) * 2 + 5;
17 ll gcd(ll a, ll b)
18 {
19     return b ? gcd(b, a % b) : a;
20 }
21 ll lcm(ll a, ll b)
22 {
23     return a * b / gcd(a, b);
24 }
25 ll d, k, tem;
26 signed main()
27 {
28     ll a, b; cin >> a >> b;
29     ll c = abs(a - b);
30     if (a < b) swap(a, b);
31     vector<ll> div;
32     for (ll i = 1; i * i <= c; i++) {     //筛出a-b的因数
33         if (c % i == 0)
34             div.push_back(i);
35         if (i * i != c)//判断是否为平方数
36             div.push_back(c / i);
37     }
38     if (a == b){                          //特判a=b的情况
39         cout << 0 << endl;
40         return 0;
41     }
42     sort(div.begin(), div.end());
43     ll ans = inf, tlcm = inf;
44     //cout << div.size() << endl;
45     d = div[0];
46     k = (d - a % d) % d;              //已知d求解k
47     tem = lcm(a + k, b + k);
48     ans = k;
49     tlcm = tem;
50     for (int i = 1; i < div.size(); i++) {    //枚举所有的因数
51         d = div[i];
52         k = (d - a % d) % d;
53         tem = lcm(a + k, b + k);
54         /*cout << i;
55         cout << "tem=" << tem << " " << "tlcm=" << tlcm << endl;
56         cout << "k=" << k << endl;*/
57         if (tem < tlcm) {  //以lcm最小为依据更新k
58             ans = k;
59             tlcm = tem;
60         }
61         else if (tem == tlcm) {   
62             ans = min(ans, k);
63         }
64         //cout << "ans=" << ans << endl;
65     }
66     cout << ans << endl;
67     return 0;
68 }

备注：由于一开始inf设置过小，循环从0开始会导致在数据过大时无法更新lcm