HDU 4901 the Romantic Hero

The Romantic herotime limit:3000msmemory limit:131072kbthis problem would be judged onHDU. Original id:4901
64-bit integer IO format: %i64d Java class name: Main There is an old country and the king fell in love with a devil. The devil always asks the king to do some crazy things. Although the king used to is wise and beloved by his people. Now he's just like a boy in love and can ' t refuse any request from the devil. Also, this devil was looking like a very cute Loli.

Wonder why is this country have such an interesting tradition? It has a very long stories, but I won ' t tell you:).

Let us continue, the party princess ' s Knight win the algorithm contest. When the devil hears on that, she decided to take some action.

But before that, there was another party arose recently, the ' Mengmengda ' Party, everyone in this party feel everything is ' Mengmengda ' and acts like a ' Mengmengda ' guy.

While they is very pleased about that, it brings many people in this kingdom troubles. So they decided to stop them.

Our hero Z*p come again, actually he was very good at algorithm contest, so he invites the leader of the ' Mengmengda ' Party Xiaod*o to compete in an algorithm contest.

As Z*p is both handsome and talkative, he had many girl friends to deal with, on the contest day, he find he had 3 dating To complete and has no time to compete, so he-let-solve the problems for him.

And the easiest problem in this contest are like that:

There is n number a_1,a_2,..., A_n on the line. You can choose the set S (A_s1,a_s2,.., A_sk) and T (A_t1,a_t2,..., a_tm). Each element in S should is at the left of every element in T. (Si < TJ for all i,j). S and T shouldn ' t be empty.

And what we want are the bitwise XOR of each element in S are equal to the bitwise and of each element in T.

How many ways is there to choose such? You should output the result modulo 10^9+7.
InputThe first line contains an integer T, denoting the number of the the test cases.
For each test case, the first line contains a integers n.
The next line contains n integers a_1,a_2,..., a_n which is separated by a single space.

n<=103, 0 <= a_i <1024, t<=20.OutputFor each test case, the output of the result in one line.Sample Input

231 2 341 2 3 3

Sample Output

1 4

SourceMulti-university Training Contest 4 Problem solving: Dynamic programming because of the existence of duplication, it is necessary to force the selection of the partition of the element, so there is dp[i][j][0] to represent the first I of the XOR and is J, and the number of options of the first I do not select Dp[i][j ][1] means the first I XOR and the number of schemes required for the first I (Dp2[i][j]) means that starting from I, followed by random selection of the number of solutions and J the answer is \[\sum_{i=1}^{n-1}\sum_{j=0}^{1023}dp[i][j][1]* Dp2[i+1][j]\]

1#include <bits/stdc++.h>2 usingLL =Long Long;3 using namespacestd;4 Const intMAXN =1010;5 ConstLL mod = 1e9 +7;6LL dp[maxn][1024x768][2],dp2[maxn][1024x768];7 intN,A[MAXN];8 intMain () {9     intKase;Tenscanf"%d",&Kase); One      while(kase--){ AMemset (DP,0,sizeofDP); -memset (DP2,0,sizeofDP2); -scanf"%d",&n); the          for(inti =1; I <= N; ++i) scanf ("%d", A +i); -dp[1][a[1]][1] =1; -          for(inti =2; I <= N; ++i) { -dp[i][a[i]][1]++; +              for(intj =0; J <1024x768; ++j) { -dp[i][j][0] + = dp[i-1][j][0] + dp[i-1][j][1]; +dp[i][j^a[i]][1] + = dp[i-1][j][0] + dp[i-1][j][1]; Adp[i][j][0] %=MoD; atdp[i][j^a[i]][1] %=MoD; -             } -         } -Dp2[n][a[n]] =1; -          for(inti = n1; i >0; --i) { -dp2[i][a[i]]++; in              for(intj =0; J <1024x768; ++j) { -DP2[I][J] + = dp2[i+1][j]; toDp2[i][j&a[i]] + = dp2[i+1][j]; +DP2[I][J]%=MoD; -Dp2[i][j&a[i]]%=MoD; the             } *         } $LL ret =0;Panax Notoginseng         for(inti =1; I < n; ++i) -          for(intj =0; J <1024x768; ++j) theRET = (ret + dp[i][j][1]*dp2[i+1][J])%MoD; +printf"%i64d\n", ret); A     } the     return 0; +}

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HDU 4901 the Romantic Hero