• [CF453B]Little Pony and Harmony Chest--状压dp


    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:

    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 integers a1, 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.

    Examples
    input
    Copy
    5
    1 1 1 1 1
    output
    Copy
    1 1 1 1 1 
    input
    Copy
    5
    1 6 4 2 8
    output
    Copy
    1 5 3 1 8 

    题解: ....(赶紧补)

     1 #include<bits/stdc++.h>
     2 using namespace std;
     3 const int p[]={2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53};
     4 const int num=(1<<16)-1;
     5 int dp[107][num+7];
     6 int choice[107][num+7];
     7 int a[107],b[107],status[67];
     8 int n;
     9 int cal(int x) {
    10     int sum=0;
    11     for (int i=0;i<16;i++)
    12         if (x%p[i]==0)
    13             sum+=(1<<i);
    14     return sum;
    15 }
    16 int main()
    17 {
    18     cin>>n;
    19     for(int i=1;i<=n;i++)  cin>>a[i];
    20     for(int i=1;i<59;i++)  status[i]=cal(i);
    21     memset(dp,0x3f,sizeof(dp));
    22     dp[0][0]=0;
    23     for(int i=1;i<=n;i++) 
    24         for(int j=0;j<=num;j++)  // num 应该等于号
    25             for(int k=1;k<59;k++) 
    26                 if((j|status[k])==j) {
    27                     if(dp[i][j]>dp[i-1][j-status[k]]+abs(k-a[i])) {
    28                         dp[i][j]=dp[i-1][j-status[k]]+abs(k-a[i]);
    29                         choice[i][j]=k;
    30                     }
    31                 }
    32     int ans=0x3f3f3f3f, k;
    33     for(int i=0;i<=num;i++) // num 这也是啊
    34         if(dp[n][i]<ans) {
    35             ans=dp[n][i];
    36             k=i;
    37         }
    38     for (int i=n;i>=1;i--) {
    39         b[i]=choice[i][k];
    40         k-=status[b[i]];
    41     }
    42     for (int i=1;i<n;i++)
    43         printf("%d ", b[i]);
    44     printf("%d
    ",b[n]);
    45     return 0;
    46 }

     

  • 相关阅读:
    自学Java0711
    自学Java0710
    自学Java0709
    自学Java0708
    Leetcode刷题集
    网站收集
    674. 最长连续递增序列『简单』
    680. 验证回文字符串 Ⅱ『简单』
    686. 重复叠加字符串匹配『简单』
    693. 交替位二进制数『简单』
  • 原文地址:https://www.cnblogs.com/xidian-mao/p/11481370.html
Copyright © 2020-2023  润新知