Codeforces Round #554 (Div. 2) C. Neko does Maths （数论 GCD(a,b) = GCD(a,b-a)）

Portal

• the meaning of problems

　　Given two positive integers a, b;

　　Solving k, such that LCM (a + k, b + k) a minimum, if there is a plurality of k such that the LCM () minimum, the minimum output of the k;

• ideas

After a lapse of a long time, again do this problem

Sure enough, one can know the new ponder ❤

Important points

GCD(a,b)=GCD(a,b-a)=GCD(b,b-a) (b>a)

prove:

Provided GCD (a, b) = c

Is a% c = 0, b% c = 0, (ba)% c = 0

Therefore GCD (a, ba) = c

得GCD(a,b)=GCD(a,b-a)

A thought:

gcd (a + k, ba) must be (ba) Factor

Therefore, gcd (a + k, b + k) is (ba) a factor, so we enumerate (ba) a factor (the factor referred i)

Such that (a + k) i is a multiple of

Solutions of the k, and then determines whether or not meet the minimum lcm

Note that only enumerated i (a + k) and a divisor of (b + k) is not necessarily the greatest common divisor gcd

Both divisor obtained common multiple common multiple = a * b / divisor

If it is a common multiple of the greatest common divisor of both must be minimized,

Here there is no discrimination whether it is the common denominator but simply get the number of convention, then get is a common multiple

In all common multiple of the least common multiple of the smallest

It does not affect the answer to the solution of the least common multiple

E.g:

12 30

12 is 30
A + K = 13 is B + K = 31 is divisor i = 1 common multiples = 403 K =. 1
A + K = 18 is B + K = 36 divisor i = 18 common multiples = 36 K =. 6
A + K = 14 B + k = 32 divisor i = 2 common multiple = 224 K = 2
A + K = 18 is B + K = 36 divisor i =. 9 common multiples = 72 K =. 6
A + K = 15 B + K = 33 is divisor i = common multiple of K 3 = 165 3 =
A + B + 18 is K = 36 K = i = 6 well divisor multiples = 108 k = 6

36 minimum number of convention, when k = 6

• Code

 1 //#include<bits/stdc++.h>
 2 //using namespace std;
 3 //#define ll long long
 4 //#define inf 0x3f3f3f3f
 5 //ll ans=inf;
 6 //ll a,b;
 7 //ll lcm;
 8 //
 9 //ll __lcm(ll x,ll y)
10 //{
11 //    return x*y/__gcd(x,y);
12 //}
13 //
14 //
15 //void Slove(int f)
16 //{
17 //    int x=a/f+(a%f!=0);
18 //    int y=(b-a)/f;
19 //    if(__gcd(x,y)==1)
20 //    {
21 //        if(__lcm(x*f,y*f)<lcm)
22 //        {
23 //            lcm=__lcm(x*f,y*f);
24 //            ans=x*f-a;
25 //        }
26 //        if(__lcm(x*f,y*f)==lcm)
27 //            ans=min(ans,x*f-a);
28 //    }
29 //    else
30 //    {
31 //        if(__lcm((x+1)*f,y*f)<lcm)
32 //        {
33 //            lcm=__lcm((x+1)*f,y*f);
34 //            ans=(x+1)*f-a;
35 //        }
36 //        if(__lcm((x+1)*f,y*f)==lcm)
37 //            ans=min(ans,(x+1)*f-a);
38 //    }
39 //}
40 //
41 //int main()
42 //{
43 //    cin>>a>>b;
44 //    if(b<a)
45 //        swap(a,b);
46 //    lcm=a*b/__gcd(a,b);
47 //
48 //    if(a==b)
49 //        ans=0;
50 //    else
51 //    {
52 //        for(ll i=1;i*i<=b-a;i++)
53 //        {
54 //            if((b-a)%i==0)
55 //            {
56 //                Slove(i);
57 //                Slove((b-a)/i);
58 //            }
59 //        }
60 //    }
61 //    cout<<ans<<endl;
62 //}
63 
64 
65 #include<bits/stdc++.h>
66 using namespace std;
67 #define ll long long
68 ll a,b;
69 ll ans,lcm=0x3f3f3f3f3f3f3f3f;
70 int main()
71 {
72     cin>>a>>b;
73     ll d=abs(a-b);
74     for(ll i=1;i*i<=d;i++)
75     {
76         IF (D% i == 0 ) // factor ba enumeration of i 
77          {
 78              LL% IA = K i; // to make up a multiple of K i needs + 
79              LL T = (K + a) * ( B + K) / I; // A * B / I have a common multiple of 
80              IF (T < LCM)
 81              {
 82                  LCM = T;
 83                  ANS = K;
 84              }
 85  
86              LL II = D / I;
 87              K = II % -a II;
 88              T = (K + A) * (B + K) / II;
 89              IF(t<lcm)
90             {
91                 lcm=t;
92                 ans=k;
93             }
94         }
95     }
96     cout<<ans<<endl;
97 }

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