题意：
给出a, b, c 3个数(a<=b<=c<=1e4), 每次操作可以对任意一个数减1或者+1，问使得bmoda cmodb == 0的操作数最少是多少？
思路：
从1，2a遍历a，对每个a，遍历<=2b的a的倍数作为b，c取c/b*b or c/b*b+b，
取最小值，时间复杂度O(nlogn).
// Decline is inevitable,
// Romance will last forever.
#include <bits/stdc++.h>
#include <iostream>
#include <cmath>
#include <cstring>
#include <string>
#include <cstdio>
#include <algorithm>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <deque>
#include <vector>
using namespace std;
#define mst(a, x) memset(a, x, sizeof(a))
#define INF 0x3f3f3f3f
//#define mp make_pair
#define pii pair<int,int>
#define fi first
#define se second
#define ll long long
//#define int long long
const int maxn = 2e5 + 10;
const int maxm = 1e3 + 10;
const int P = 1e9 + 7;
int n, m;
int a, b, c;
void solve() {
cin >> a >> b >> c;
if(c % b == 0 && b % a == 0) {
cout << 0 << endl;
cout << a << ' ' << b << ' ' << c << endl;
return;
}
int tmp = INF;
int ans[3];
for(int i = 1; i <= 2*a; i++) {
for(int j = i; j <= 2*b; j += i) {
int c1 = c/j * j;
int c2 = (c/j+1) * j;
int ans1 = abs(a-i) + abs(b-j) + abs(c1-c);
int ans2 = abs(a-i) + abs(b-j) + abs(c2-c);
if(ans1 < tmp) {
ans[0] = i; ans[1] = j; ans[2] = c1;
tmp = ans1;
}
if(ans2 < tmp) {
ans[0] = i; ans[1] = j; ans[2] = c2;
tmp = ans2;
}
}
}
cout << tmp << endl;
for(int i = 0; i < 3; i++)
cout << ans[i] << ' ';
cout << endl;
}
signed main() {
ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
// int T; scanf("%d", &T); while(T--)
int T; cin >> T; while(T--)
solve();
return 0;
}