Codeforces Round #259 (Div. 1) -- Little Pony and Harmony Chest,

Codeforces Round #259 (Div. 1) -- Little Pony and Harmony Chest,

Question connection

• Question:
  Calculate a sequence bi for n Integers of ai, so that any two numbers in the B sequence are mutually dependent, and sigma (abs (ai-bi) is the smallest, and output any B sequence.
  (1 digit ≤ DigitNMemory ≤ memory 100) (1 memory ≤ memoryAILimit ≤ Limit 30)
  
• Analysis:
  First of all, it is clear that the scope of question B is not limited .... For this reason, wa has been performed many times, but it can be inferred that B must be less than or equal to 60.
  Any two numbers are mutually qualitative. That is to say, if you know the quality factor of all the previous numbers for a newly added bi, It is a satisfying situation as long as the current number does not have this quality factor. There are less than 20 Prime numbers within 60, so you can directly press DP. DP [I] [j] indicates the number of I processed. The prime factor is j (in binary format)

const int MAXN = 60;const int SIZE = 16;int ipt[110];int dp[110][1 << SIZE];int p[110][1 << SIZE];int x[110][1 << SIZE];int to[MAXN + 1];int prime[] = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53};map<int, int> mp;void init(){    REP(i, SIZE)        mp[prime[i]] = 1 << i;    FE(ii, 0, MAXN)    {        int n = ii, ret = 0, i;        for (i = 0; i < SIZE && prime[i] * prime[i] <= n; i++)            if (n % prime[i] == 0)            {                while (n % prime[i] == 0)                    n /= prime[i];                ret += 1 << i;            }        if (n > 1)            ret += mp[n];        to[ii] = ret;    }}int n;void dfs(int idx, int id){    if (idx >= 2)        dfs(idx - 1, p[idx][id]);    printf("%d%c", x[idx][id], idx == n ? '\n' : ' ');}int main (){    int all = 1 << SIZE;    init();    while (~RI(n))    {        FE(i, 1, n)            RI(ipt[i]);        CLR(dp, INF); dp[0][0] = 0;        FE(i, 0, n - 1)        {            FF(j, 0, all)            {                FE(jj, 0, MAXN)                {                    int v = to[jj];                    if (j & v)                        continue;                    int val = dp[i][j] + abs(jj - ipt[i + 1]);                    int& nxt = dp[i + 1][j | v];                    if (val < nxt)                    {                        nxt = val;                        p[i + 1][j | v] = j;                        x[i + 1][j | v] = jj;                    }                }            }        }        int ans = INF, id = 0;        REP(j, all)        {            if (dp[n][j] < ans)            {                ans = dp[n][j];                id = j;            }        }        dfs(n, id);    }    return 0;}