C. Neko does Maths(数论 二进制枚举因数)
题目链接：https://codeforces.com/contest/1152/problem/C
题目大意：给你a和b，然后让你找到一个k,使得a+k和b+k的lcm.
学习网址：https://blog.csdn.net/yopilipala/article/details/89517933
具体思路：
AC代码：
1 #include<bits/stdc++.h>
2 using namespace std;
3 # define ll long long
4 # define inf 0x3f3f3f3f
5 const int maxn = 2e5+100;
6 vector<ll>sto;
7 void init(ll t)
8 {
9 ll tmp=t;
10 for(ll i=2; i*i<=tmp; i++)
11 {
12 while(t%i==0)
13 {
14 t/=i;
15 sto.push_back(i);
16 }
17 }
18 if(t!=1)
19 sto.push_back(t);
20 }
21 ll cal(int t)
22 {
23 ll ans=1ll;
24 int pos=0;
25 while(t)
26 {
27 if(t&1)
28 {
29 ans=ans*sto[pos];
30 }
31 pos++;
32 t>>=1;
33 }
34 return ans;
35 }
36 int main()
37 {
38 ll a,b;
39 scanf("%lld %lld",&a,&b);
40 if(a==b)
41 {
42 printf("0\n");
43 return 0;
44 }
45 if(a>b)
46 swap(a,b);
47 init(b-a);
48 int maxstate=(1<<(sto.size()))-1;
49 ll minn=(1ll<<60);
50 ll ans=(1ll<<60);
51 for(int i=1; i<=maxstate; i++)
52 {
53 ll tmp=cal(i);
54 ll t1=(a/tmp+1)*tmp;//注意是先除
55 ll t2=(b/tmp+1)*tmp;
56 ll w=t1*t2/__gcd(t1,t2);
57 if(w<=minn)
58 {
59 if(w<minn)
60 {
61 minn=w;
62 ans=t1;
63 }
64 else if(w==minn)
65 {
66 minn=w;
67 ans=min(ans,t1);
68 }
69 }
70 }
71 ll tmp=a*b/__gcd(a,b);
72 if(tmp<=minn)
73 {
74 minn=tmp;
75 ans=a;
76 }
77 printf("%lld\n",ans-a);
78 return 0;
79 }
