hdoj4906 our happy Ending (multi-university Training Contest 4)

For a fixed-length (size = N) sequence A, if there is a subset of "positional correlation" (with an empty set) that makes the sum of all the elements of the subset K, then the sequence A is counted.

Where the sequence a of any element a[i] in [0, L] free to take the value.

Data conditions 0≤n, k≤20, 0≤L≤1E9, count results for mod = 1e9 + 7 modulo.

No matter the direct count or the consideration from the negative count can not solve the problem of heavy, can only consider DP.

Enumerates the length of the sequence I and the compressed state J, and records the number of sequence selection schemes under this condition DP[I][J].

The compressed State J indicates the selection of the collection {1, 2, ..., Min (l, K)}.

Where the first element of the set is in the state J, and only if the first bit of the binary string J is 1.

Obviously we have dp[0][0] = 1.

For an array of length p, the element that can be selected at p bit is 0,1,2,...,l

Consider P-bit selection 1,2,...,min (L, K)

Then for the any state J when the array length is p-1, the state after appending the element I to the array is:

J1 = J | ((J << i) & ((1 << k)-1)) | (1 << (i-1))

The three parts represent the original state, the state of each element in the original state and the increment of the element I, I itself.

So dp[p][j1] + = Dp[p-1][j].

For P-bit selection 0 or greater than min (L, k), J1 = J.

Therefore there is dp[p][j] = (L-min (l, k)) * Dp[p-1][j].

We finally only need to dp[n][j], where J K bit is 1 cumulative.

Because each element of the sequence is non-negative, we finally only care about those sets that are present and a subset of K,

And those elements greater than k are necessarily not elements that make up a subset, only those elements that are less than k are useful for state transitions.

So we can represent states like this.

1#include <cstdio>2#include <cstring>3#include <algorithm>4#include <map>5#include <string>6#include <vector>7#include <Set>8#include <cmath>9#include <ctime>Ten using namespacestd; One #defineLson (U << 1) A #defineRson (U << 1 | 1) - typedef __int64 LL; - Const intMAXN = 1e6 +Ten; the Const intMAXM =1050; - Constll mod = 1e9 +7; -  - intN, l, K; +ll dp[(1<< +) +Ten]; - intMain () { +     intT; Ascanf"%d", &T); at      while(t--){ -scanf"%d%d%d", &n, &k, &l); -ll d = ABS (L-k); -L =min (l, k); -         ints = (1<< k)-1; -Memset (DP,0,sizeofDP); indp[0] =1; -          while(n--){ to              for(intj = S; J >=0; j--){ +ll tem =Dp[j]; -                 if(!tem)Continue; the                  for(intp =1; P <= l; p++){ *                     intNEX = (1<< (P-1)) | J | (j << P) &s); $Dp[nex] = (Dp[nex] + tem)%MoD;Panax Notoginseng                 } -DP[J] = (TEM * (1+ D))%MoD; the             } +         } All ans =0; the          for(inti = s; I >=0; i--) +             if(I & (1<< (K-1))) -Ans = (ans + dp[i])%MoD; $printf"%i64d\n", ans); $     } -     return 0; -}

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hdoj4906 our happy Ending (multi-university Training Contest 4)