Codeforces Round #259 (Div. 2) D. Little Pony and Harmony Chest 状压DP

时间:2023-03-09 02:10:23
Codeforces Round #259 (Div. 2) D. Little Pony and Harmony Chest  状压DP

D. Little Pony and Harmony Chest

Princess Twilight went to Celestia and Luna's old castle to research the chest from the Elements of Harmony.

A sequence of positive integers bi is harmony if and only if for every two elements of the sequence their greatest common divisor equals 1. According to an ancient book, the key of the chest is a harmony sequence bi which minimizes the following expression:

Codeforces Round #259 (Div. 2) D. Little Pony and Harmony Chest  状压DP

You are given sequence ai, help Princess Twilight to find the key.

Input

The first line contains an integer n (1 ≤ n ≤ 100) — the number of elements of the sequences a and b. The next line contains n integersa1, a2, ..., an (1 ≤ ai ≤ 30).

Output

Output the key — sequence bi that minimizes the sum described above. If there are multiple optimal sequences, you can output any of them.

题意:

给你一个长度不超过100的数组A, 求数组B使得对应元素差值的和最小,且B数组两两元素互质。

很容易看出来,B[i]最大值不会超过58, 59也可以但是用59的话为什么不用1呢。

用mask[i]表示i的质因子,比如mask[6] = 3

dp[i][mask[j]|k] = min(dp[i-1][j] + abs(A[i] - k)) 表示前i个数中已经选了质因子的状态为mask[j]|k, 转移的时候要保证mask[j] & k == 0,

然后记录转移的过程(为了输出B),然后就完了