HDU4570 ---- Multi-bit Trie ---- simple DP

Question meaning:

N count

You can divide it into multiple segments. The length of each segment cannot exceed 20.

Yes, sum (ai * (2 ^ bi) is the smallest. ai is the first number in each segment, and bi is the length.

Solution:

Set dp [I] = min (dp [I], dp [j] + a [I] * 2 ^ (j-I), 1 <= I <= n, I + 1 <= j <= min (I + 20, n + 1)

Dp [I] indicates the value starting with I

Finally, we will compare dp [1] with a comparison into n segments.

There are a lot of code on the Internet. I wrote an iteration, which is faster than the memory-based search.

 

#include<cstdio>   #include<cstring>   #include<cmath>   #include<algorithm>   #include<iostream>   using namespace std;    #define ULL long long     const int maxn = 65;    ULL dp[maxn];  ULL a[maxn];  int n;    int main()  {      int t;      scanf("%d",&t);      while(t--)      {          scanf("%d",&n);          ULL ans=0;          for(int i=1;i<=n;i++)          {              cin>>a[i];              ans += a[i]*2;          }            dp[n] = a[n]*2;          dp[n+1] = 0;          for(int i=n-1;i>=1;i--)          {              dp[i] = (1ULL<<62-1);              for(int j=i+1;j<=min(i+20,n+1);j++)              {                  dp[i] = min(dp[i],dp[j]+a[i]*(1<<(j-i)));              }          }            cout<<min(ans,dp[1])<<endl;      }      return 0;  }