传送门

题意：

给定$A,B,C$，找到一个三元组$(a,b,c)$，使得$a$是$b$的因子，$b$是$c$的因子，且$abs(A-a)+abs(B-b)+abs(C-c)$最小。
$A,B,C\le 1e4$

思路：

注意到$a$是$b$的因子，$b$是$c$的因子，所以考虑枚举$a$，那么枚举$b$就是$k\ast a$，枚举$c$就是$t\ast b$，枚举上限为$2\ast c$，复杂度$O(nlo{g}^{2}n)$。
当然还可以枚举$b$，之后枚举$b$的因子，找到离$a$最近的那个，让后再枚举$b$的倍数，找到离$c$最近的那个即可。

$O(nlo{g}^n)$

// Problem: D. Three Integers
// Contest: Codeforces - Codeforces Round #624 (Div. 3)
// URL: https://codeforces.com/contest/1311/problem/D
// Memory Limit: 256 MB
// Time Limit: 2000 ms
// 
// Powered by CP Editor (https://cpeditor.org)#pragma GCC optimize(2)
#include<cstdio>
#include<iostream>
#include<string>
#include<cstring>
#include<map>
#include<cmath>
#include<cctype>
#include<vector>
#include<set>
#include<queue>
#include<algorithm>
#include<sstream>
#define X first
#define Y second
#define L (u<<1)
#define R (u<<1|1)
#define Mid (tr[u].l+tr[u].r>>1)
#define Len(u) (tr[u].r-tr[u].l+1)
#define pb push_back
#define mk make_pair
using namespace std;typedef long long LL;
typedef pair<int,int> PII;const int N=200010,mod=1e9+7,INF=0x3f3f3f3f;
const double eps=1e-6;int a,b,c;int main()
{
//	ios::sync_with_stdio(false);
//	cin.tie(0);int _; scanf("%d",&_);while(_--){scanf("%d%d%d",&a,&b,&c);int a1,a2,a3,ans=INF;for(int i=1;i<=2*c;i++){for(int j=i;j<=2*c;j+=i){for(int k=j;k<=2*c;k+=j){int dis=abs(i-a)+abs(j-b)+abs(k-c);if(ans>dis) ans=dis,a1=i,a2=j,a3=k;}}}printf("%d\n%d %d %d\n",ans,a1,a2,a3);}	return 0;
}
/**/

$O(nlogn)$

// Problem: D. Three Integers
// Contest: Codeforces - Codeforces Round #624 (Div. 3)
// URL: https://codeforces.com/contest/1311/problem/D
// Memory Limit: 256 MB
// Time Limit: 2000 ms
// 
// Powered by CP Editor (https://cpeditor.org)#pragma GCC optimize(2)
#include<cstdio>
#include<iostream>
#include<string>
#include<cstring>
#include<map>
#include<cmath>
#include<cctype>
#include<vector>
#include<set>
#include<queue>
#include<algorithm>
#include<sstream>
#define X first
#define Y second
#define L (u<<1)
#define R (u<<1|1)
#define Mid (tr[u].l+tr[u].r>>1)
#define Len(u) (tr[u].r-tr[u].l+1)
#define pb push_back
#define mk make_pair
using namespace std;typedef long long LL;
typedef pair<int,int> PII;const int N=200010,mod=1e9+7,INF=0x3f3f3f3f;
const double eps=1e-6;int a,b,c;
vector<int>d[N];int main()
{
//	ios::sync_with_stdio(false);
//	cin.tie(0);for(int i=1;i<N;i++)for(int j=i;j<N;j+=i)d[j].pb(i);int _; scanf("%d",&_);while(_--){scanf("%d%d%d",&a,&b,&c);int a1,a2,a3,ans=INF;for(int i=1;i<=2*c;i++){int x,y,z,cnt=0,mi=INF;for(auto xx:d[i]) if(abs(xx-a)<mi) mi=abs(xx-a),x=xx;cnt+=mi; mi=INF;for(int j=i;j<=2*c;j+=i) if(abs(j-c)<mi) mi=abs(j-c),y=j;cnt+=mi;cnt+=abs(i-b);if(cnt<ans) ans=cnt,a1=x,a2=i,a3=y;}printf("%d\n%d %d %d\n",ans,a1,a2,a3);}	return 0;
}
/**/