"ZOJ 5689" Deque and Balls (normal DP)

Test Instructions: give a sequence, in order one by one into the double-ended queue (can be placed in the head can also be placed in the tail), a queue of beautiful index is the number of a[i]>a[i+1 in the series , to seek the expectation of beauty index *2^n the value.

problem-solving ideas: for convenience, we refer to a[i]>a[i+1] as the case of D .

By test instructions can know that the last obtained sequence has a total of 2^ (n-1) , set the sum of the number of D in all sequences of the ans= , the last is to ask sum/2^ (n-1) *2^n = 2*sum

For a[j " sum= original case sum *2+a[j] The resulting d -(A[i]=a[j]) The case of

If a[j] is adjacent to A[i] , then a[i+1 ... J-1] can only be placed in the A[i] the opposite end, then appears D the sequence of the conditions altogether has 2^ (i-2) a

So the D case of A[j] has altogether 1+∑2^ (i-2) (2<=I<J)

A[I]=A[J] The situation is similar to the idea of J .

1#include <bits/stdc++.h>2 using namespacestd;3 #definell Long Long4 Const intn=1e5+Ten;5 Constll mod=1e9+7;6 intA;7 ll F[n], Sum[n], dp[n], p[n];8 intMain () {9f[0]=1, f[1]=1, sum[0]=1, sum[1]=2;Ten      for(intI=2; i<n; i++) F[i] = (f[i-1]*2)%mod, sum[i] = (sum[i-1]+f[i])%MoD; One      A     intt, N; -scanf"%d", &t); -      while(t--){ theMemset (P,0,sizeof(P)); -scanf"%d", &n); -dp[0] =0, dp[1] =0; -          for(intI=1; i<=n; i++){ +scanf"%d", &a); -             if(i>1) Dp[i] = ((dp[i-1]*2)%mod + sum[i-2]-p[a] + MoD)%MoD; +P[a] = (p[a]+f[i-1])%MoD; A         } atprintf"%lld\n", (dp[n]*2)%MoD); -     } -     return 0; -}

"ZOJ 5689" Deque and Balls (normal DP)