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 }