# C. Neko does Maths

time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
Neko loves divisors. During the latest number theory lesson, he got an interesting exercise from his math teacher.

Neko has two integers a and b. His goal is to find a non-negative integer k such that the least common multiple of a+k and b+k is the smallest possible. If there are multiple optimal integers k, he needs to choose the smallest one.

Given his mathematical talent, Neko had no trouble getting Wrong Answer on this problem. Can you help him solve it?

Input
The only line contains two integers a and b (1≤a,b≤109).

Output
Print the smallest non-negative integer k (k≥0) such that the lowest common multiple of a+k and b+k is the smallest possible.

If there are many possible integers k giving the same value of the least common multiple, print the smallest one.

Examples
inputCopy
6 10
outputCopy
2
inputCopy
21 31
outputCopy
9
inputCopy
5 10
outputCopy
0
Note
In the first test, one should choose k=2, as the least common multiple of 6+2 and 10+2 is 24, which is the smallest least common multiple possible.

## 解法lcm(x,y)=x*y/gcd(x,y)，那么求gcd(x+k,y+k)的最大就好了，而知道gcd(x,y)=gcd(y-x,x)，因为如果gcd(x,y)=c，那么，x%c=0，y%c=0，(y-x)%c=0，则，求gcd(y-x,x+k)，求出所有的y-x的因子，然后全部都拿来算一下，现在要知道因子，求对应的k值，可知k+x是因子的倍数，则k=因子-（因子%x）就可以了，然后每一次算一下lcm，得出最大的保留就好了

``````#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstdio>
#include<cmath>
#include<queue>
#include<vector>
#include<stack>
#include<map>
#define sf scanf
#define scf(x) scanf("%lld",&x)
#define scff(x,y) scanf("%lld%lld",&x,&y)
#define scfff(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define vi vector<int>
#define mp make_pair
#define pf printf
#define prf(x) printf("%d\n",x)
#define mm(x,b) memset((x),(b),sizeof(x))
#define rep(i,a,n) for (ll i=a;i<n;i++)
#define per(i,a,n) for (int i=a;i>=n;i--)
typedef long long ll;
using namespace std;
const ll mod=1e9+7;
const double eps=1e-6;
const double pi=acos(-1.0);
const int inf=0x7fffffff;
const int N=1e7+7;
ll gcd(ll x,ll y)
{
return y==0?x:gcd(y,x%y);
}
ll lcm(ll x,ll y)
{
return x*y/(gcd(x,y));
}
vector<int> v;
int main()
{
ll x,y;scff(x,y);
if(x>y) swap(x,y);
ll ans=0,maxn=lcm(x,y),ss=y-x;
for(ll i=1;i*i<=ss;i++)
{
if(ss%i==0)
{
v.push_back(i);
if(i*i!=ss)
v.push_back(ss/i);
}
}
rep(i,0,v.size() )
{
ll t=0;
if(x%v[i]!=0)
t=v[i]-x%v[i];
ll now=lcm(x+t,y+t);
if(now<maxn)
{
maxn=now;
ans=t;
}
}
cout<<ans;
return 0;
}
``````

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11 q. 0.074 s.