height accep fin -a can stop ott from include

The Romantic Hero

Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1128 Accepted Submission(s): 469


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

2
3
1 2 3
4
1 2 3 3

Sample Output

1 
4

題意是 給n個數.

前面取隨意個數 組成集合S 取在全部S中元素後面的隨意個元素 T ,S 和T 不為空集;

計算能讓S集合全部元素的異或(^)後得到的數等於 T集合全部元素與(&)後得到的數. 這樣取兩個集合,不同的取法有多少個.

data[ i ][ j ]表示在i號元素曾經,包含i, 能通過^運算得到j的方法數

data2[ i ][ j ]表示在i號元素以後,包含i,能通過&運算得到j的方法數

data3[ i ][ j ] 表示在i號元素以後,包含i,能通過&運算得到j,且一定取了i, 的方法數.

然後,不斷遞推,就能夠了.

#include<stdio.h>
#include<string.h>
__int64 data[2000][1200],data2[2000][1200],data3[2000][1200];
int main()
{
    int t,n,i,a[2000],j;
    __int64 ans;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        for(i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
        }
        memset(data,0,sizeof(data));
        memset(data2,0,sizeof(data2));
        memset(data3,0,sizeof(data3));
        for(i=1;i<=n;i++)
        {
            if(i!=1)
            {
                for(j=0;j<1024;j++)
                {
                    if(data[i-1][j])
                    {
                        data[i][j^a[i]]+=data[i-1][j];//取了當前這個
                        data[i][j^a[i]]%=(1000000000+7);
                        data[i][j]+=data[i-1][j];//沒取當前這個
                        data[i][j]%=(1000000000+7);
                    }
                }
            }
            data[i][a[i]]++;//僅僅取當前這個 前面的都沒取
            data[i][a[i]]%=(1000000000+7);
        }
        ans=0;
        for(i=n;i>=1;i--)//data2  代表所有的  data3 代表取了當前這個的,避免反復計算
        {
            data3[i][a[i]]++;//僅僅取了當前這個
            data3[i][a[i]]%=(1000000000+7);
            data2[i][a[i]]++;
            data2[i][a[i]]%=(1000000000+7);
            if(i!=n)
            {
                for(j=0;j<1024;j++)
                {
                    if(data2[i+1][j])
                    {
                        data2[i][j&a[i]]+=data2[i+1][j];//取了當前這個
                        data2[i][j&a[i]]%=(1000000000+7);
                        data3[i][j&a[i]]+=data2[i+1][j];
                        data3[i][j&a[i]]%=(1000000000+7);
                        data2[i][j]+=data2[i+1][j];//沒取當前這個
                        data2[i][j]%=(1000000000+7);
                    }
                }
            }
            for(j=0;j<1024;j++)
            {
                if(data3[i][j])
                {
                    ans+=data[i-1][j]*data3[i][j];
                    ans=ans%(1000000000+7);
                }
            }
        }
        printf("%I64d\n",ans);
    }
    return 0;
}

hdu 4901 The Romantic Hero 計數dp,位計算