hdu 4901 The Romantic Hero（计数dp）2014多校训练第4场1005

The Romantic Hero

                                                                              Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)

Problem Description

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 be wise and beloved by his people. Now he is just like a boy in love and can’t refuse any request from the devil. Also, this devil is looking like a very cute Loli.

You may wonder why this country has such an interesting tradition? It has a very long story, but I won't tell you :).

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

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

While they are 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 is 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 has many girl friends to deal with, on the contest day, he find he has 3 dating to complete and have no time to compete, so he let you to solve the problems for him.

And the easiest problem in this contest is like that:

There is n number a_1,a_2,...,a_n on the line. You can choose two set S(a_s1,a_s2,..,a_sk) and T(a_t1,a_t2,...,a_tm). Each element in S should be 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 is the bitwise XOR of each element in S is equal to the bitwise AND of each element in T.

How many ways are there to choose such two sets? You should output the result modulo 10^9+7.

 

Input

The first line contains an integer T, denoting the number of 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 are separated by a single space.

n<=10^3, 0 <= a_i <1024, T<=20.

 

Output

For each test case, output the result in one line.

 

Sample Input

231 2 341 2 3 3

 

Sample Output

1 4

 

题意：给出n个数，构造两个序列，使得第一个序列里面所有元素的异或值等于第二个序列里面所有元素的AND（&）值，并且第一个序列里所有元素的下标都小于第二个序列里所有元素的下标。求一共有多少种构造方法，结果对1000000007取余。

虽然比赛时就知道是dp，但是由于dp功底太弱，导致比赛时没有做出来。

分析：

dp1[i][j]：由0~i的元素异或得到j的种类数。
dp2[i][j]：由i~n-1的元素AND得到j的种类数。

dp3[i][j]：由i~n-1的元素，且一定包含a[i]，AND得到j的种类数。

求出这些，最后把dp1[i][j]*dp3[i+1][j]求和就得到答案了！
这里多用了一个数组dp3，而不是直接用dp2，是为了防止重复计数。

#include<cstdio>#include<cstring>typedef __int64 LL;#define mod 1000000007const int MAXN = 1002;const int MAXA = 1025;int dp1[MAXN][MAXA], dp2[MAXN][MAXA], dp3[MAXN][MAXA];int a[MAXN];int main(){    int T, n, i, j, t;    scanf("%d",&T);    while(T--) {        scanf("%d",&n);        for(i = 0; i < n; i++)            scanf("%d",&a[i]);        memset(dp1, 0, sizeof(dp1));        memset(dp2, 0, sizeof(dp2));        memset(dp3, 0, sizeof(dp3));        dp1[0][a[0]] = 1;        for(i = 1; i < n - 1; i++) {            dp1[i][a[i]]++; //单独一个元素构成一个集合            for(j = 0; j < MAXA; j++) {                if(dp1[i-1][j]) {                    dp1[i][j] += dp1[i-1][j]; //不添加第i个元素进行异或,继承之前算好的                    dp1[i][j] %= mod;                    t = j ^ a[i];  //添加第i个元素进行异或                    dp1[i][t] += dp1[i-1][j];                    dp1[i][t] %= mod;                }            }        }        dp2[n-1][a[n-1]] = 1;        dp3[n-1][a[n-1]] = 1;        for(i = n-2; i > 0; i--) {            dp2[i][a[i]]++;            dp3[i][a[i]]++;   //单独一个元素构成一个集合            for(j = 0; j < MAXA; j++) {                if(dp2[i+1][j]) {                    dp2[i][j] += dp2[i+1][j];  //不添加第i个元素进行按位与                    dp2[i][j] %= mod;                    t = j & a[i]; //添加第i个元素进行按位与                    dp2[i][t] += dp2[i+1][j];                    dp2[i][t] %= mod;                    dp3[i][t] += dp2[i+1][j]; //添加第i个元素进行按位与                    dp3[i][t] %= mod;                }            }        }        int ans = 0;        for(i = 0; i < n - 1; i++) {            for(j = 0; j < MAXA; j++) {                if(dp1[i][j] && dp3[i+1][j]) {                    ans += (LL(dp1[i][j]) * dp3[i+1][j] % mod);                    ans %= mod;                }            }        }        printf("%d\n", ans);    }    return 0;}