HDU4901 The Romantic Hero 计数DP

2014多校4的1005

题目：http://acm.hdu.edu.cn/showproblem.php?pid=4901

The Romantic Hero

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others) Total Submission(s): 393    Accepted Submission(s): 150

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

 

Author

WJMZBMR

 

Source

2014 Multi-University Training Contest 4

 

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题意：

    给一列数，让你选两个集合，A集合所有元素下标小于B集合所有元素下标，A集合的所有元素异或等于B集合所有元素AND，两个集合都非空，求集合元素有多少种选法，MOD 10^9+7。0<=元素大小<1024，最多1000个元素。

 

题解：

    DP！（虽然我一开始就想到DP，不过我DP功力太差，比赛中实在做不出，结束后看了http://www.cnblogs.com/avema/p/3881466.html的超快题解（虽然没写题解只有代码）才学会的）

    官方题解居然是这样的“水题，大家都会做吧？”我都怕了！虽然很多人过了，好像的确很水的样子……

    言归正传，我来讲一下这个DP。

    f[i][j]：由0~i的元素异或得到j的种类数。

    h[i][j]：由i~n-1的元素AND得到j的种类数。

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

求出这些，最后把f[i][j]*g[i+1][j]求和就得到答案了！

这里用g而不用h，防止重复计数，非常高端。我写的时候也一直在考虑防止重复计数，结果写出来4重循环，我自己都怕，看来是我对DP的理解不够深。这个只用两重循环，快得飞起来。具体实现看代码吧，还算比较简单易懂。

代码：

#include<stdio.h>
#include<math.h>
#include<string.h>
#include<math.h>

#define ll __int64
#define usint unsigned int
#define mz(array) memset(array, 0, sizeof(array))
#define RE  freopen("1.in","r",stdin)
#define WE  freopen("mask.txt","w",stdout)

#define maxa 1024
#define maxn 1000
#define C 1000000007

int f[maxn-1][maxa],g[maxn][maxa],h[maxn][maxa];

int main() {

    int a[maxn],T,n;
    short i,j,t;
    int ans;
    scanf("%d",&T);
    while(T--) {
        scanf("%d",&n);
        for(i=0; i<n; i++)
            scanf("%d",&a[i]);
        mz(f);
        mz(g);
        mz(h);
        f[0][a[0]]=1;
        for(i=1; i<n-1; i++) {
            f[i][a[i]]++;///单独一个元素的集合的情况
            for(j=0; j<maxa; j++) {
                if(f[i-1][j]) {
                    f[i][j]+=f[i-1][j];///继承之前算好的情况（就是 不包括当前元素的情况）
                    f[i][j]%=C;
                    t=j^a[i];
                    f[i][t]+=f[i-1][j];///由前一次的情况异或当前元素得到的情况（包括当前元素的情况）
                    f[i][t]%=C;
                }

            }
        }

        g[n-1][a[n-1]]=1;
        h[n-1][a[n-1]]=1;
        for(i=n-2; i>0; i--) {
            g[i][a[i]]++;
            h[i][a[i]]++;
            for(j=0; j<maxa; j++) {
                if(h[i+1][j]) {
                    h[i][j]+=h[i+1][j];
                    h[i][j]%=C;
                    t=j&a[i];
                    h[i][t]+=h[i+1][j];
                    h[i][t]%=C;

                    g[i][t]+=h[i+1][j];///包括当前元素的情况（g没有不包括当前元素的情况）
                    g[i][t]%=C;
                }

            }
        }
        ans=0;
        for(i=0; i<n-1; i++)
            for(j=0; j<maxa; j++) {
                if(f[i][j]&&g[i+1][j]) {
                    ans+=(((ll)f[i][j])*(g[i+1][j])%C);
                    ans%=C;
                }
            }
        printf("%d\n",ans);
    }
    return 0;
}

这次来个C的代码，酷不酷炫（其实没有类都可以换成C的吧