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D. Three Integers
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
You are given three integers a≤b≤ca≤b≤c.
In one move, you can add +1+1 or −1−1 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≤CA≤B≤C such that BB is divisible by AA and CC is divisible by BB.
You have to answer tt independent test cases.
Input
The first line of the input contains one integer tt (1≤t≤1001≤t≤100) — the number of test cases.
The next tt lines describe test cases. Each test case is given on a separate line as three space-separated integers a,ba,b and cc (1≤a≤b≤c≤1041≤a≤b≤c≤104).
Output
For each test case, print the answer. In the first line print resres — the minimum number of operations you have to perform to obtain three integers A≤B≤CA≤B≤C such that BB is divisible by AA and CC is divisible by BB. On the second line print any suitable triple A,BA,B and CC.
Example
input
Copy
81 2 3123 321 4565 10 1515 18 21100 100 1011 22 293 19 386 30 46
output
Copy
11 1 3102114 228 45644 8 16618 18 181100 100 10071 22 2221 19 3886 24 48
【题意】
对于给定的三个数a,b,c，可以进行+1或者-1操作，最后要满足b=k1*a,c=k2*b=k1*k2*a，求满足这个要求所需要的最小操作数，就是枚举的数与a,b,c的差值最小。
因为数据给了1e4，所以直接暴力三层循环枚举，按照倍数枚举，最后复杂度不会是O(n^3)，a枚举到10000，b枚举到20000，一般不会超过这个。
#includeusing namespace std;const int inf=0x3f3f3f3f;int main(){ ios::sync_with_stdio(false); int t,a,b,c; int x,y,z; cin>>t; while(t--) { cin>>a>>b>>c; int ans=inf; for(int i=1;i<=10000;i++) { for(int j=1;j*i<=20000;j++) { for(int k=1;i*j*k<=20000;k++) { int sum=abs(a-i)+abs(j*i-b)+abs(i*j*k-c); if(ans>sum) { ans=sum; x=i; y=i*j; z=i*j*k; } } } } cout< <
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