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：

 #include<bits/stdc++.h>
 using namespace std;
 # define ll long long
 # define inf 0x3f3f3f3f
 const int maxn = 2e5+;
 vector<ll>sto;
 void init(ll t)
 {
     ll tmp=t;
     for(ll i=; i*i<=tmp; i++)
     {
         while(t%i==)
         {
             t/=i;
             sto.push_back(i);
         }
     }
     if(t!=)
         sto.push_back(t);
 }
 ll cal(int t)
 {
     ll ans=1ll;
     int pos=;
     while(t)
     {
         if(t&)
         {
             ans=ans*sto[pos];
         }
         pos++;
         t>>=;
     }
     return ans;
 }
 int main()
 {
     ll a,b;
     scanf("%lld %lld",&a,&b);
     if(a==b)
     {
         printf("0\n");
         return ;
     }
     if(a>b)
         swap(a,b);
     init(b-a);
     int maxstate=(<<(sto.size()))-;
     ll minn=(1ll<<);
     ll ans=(1ll<<);
     for(int i=; i<=maxstate; i++)
     {
         ll tmp=cal(i);
         ll t1=(a/tmp+)*tmp;//注意是先除
         ll t2=(b/tmp+)*tmp;
         ll w=t1*t2/__gcd(t1,t2);
         if(w<=minn)
         {
             if(w<minn)
             {
                 minn=w;
                 ans=t1;
             }
             else if(w==minn)
             {
                 minn=w;
                 ans=min(ans,t1);
             }
         }
     }
     ll tmp=a*b/__gcd(a,b);
     if(tmp<=minn)
     {
         minn=tmp;
         ans=a;
     }
     printf("%lld\n",ans-a);
     return ;
 }