Codeforces Round ＃624 (Div. 3)

Codeforces Round #624 (Div. 3)

D. Three Integers

You are given three integers a≤b≤c.
In one move, you can add +1 or −1
to any of these integers (i.e. increase or decrease any number by one). You can perform such operation any (possibly, zero) number of times, you can even perform this operation several times with one number. Note that you cannot make non-positive numbers using such operations.
You have to perform the minimum number of such operations in order to obtain three integers A≤B≤C such that B is divisible by A and C is divisible by B.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t(1≤t≤100) — the number of test cases.
The next t lines describe test cases. Each test case is given on a separate line as three space-separated integers a,b and c (1≤a≤b≤c≤104).
Output
For each test case, print the answer. In the first line print res— the minimum number of operations you have to perform to obtain three integers A≤B≤C such that B is divisible by A and C is divisible by B. On the second line print any suitable triple A,B and C.
Example
Input

8
1 2 3
123 321 456
5 10 15
15 18 21
100 100 101
1 22 29
3 19 38
6 30 46

Output

1
1 1 3
102
114 228 456
4
4 8 16
6
18 18 18
1
100 100 100
7
1 22 22
2
1 19 38
8
6 24 48

解析
题目大意就是，给你三个数a,b,c,进行+1或-1操作，使得 b%a== 0&&c%b == 0 ，找出操作次数最少的。
这个题目用普通的正面思考会很复杂，第一个想到的就是bfs，但是用bfs很容易爆内存。
简单的方法是换一种思路：从满足b%a== 0&&c%b == 0条件开始找最小的值。
ac代码

#include<bits/stdc++.h>
using namespace std;
int t,a,b,c;
int main()
{
    
	scanf("%d",&t);
	while(t--)
	{
    
		cin>>a>>b>>c;
		int v1=0,v2=0,v3=0,ans=2147483647;
		for(int i=1;i<=11000;i++)
		  for(int j=i;j<=11000;j+=i)
		    for(int k=j;k<=11000;k+=j)
		      if(abs(a-i)+abs(b-j)+abs(c-k)<ans)
		        ans=abs(a-i)+abs(b-j)+abs(c-k),v1=i,v2=j,v3=k;
		printf("%d\n",ans);
		printf("%d %d %d\n",v1,v2,v3);
	}
	return 0;
}