SRM 508(2-1000pt)

DIV2 1000pt

题意：给定整数n和r，求有多少个这样的数列，a1，a2...an，使得a1 + a2 +...+an = a1|a2|a3|...|an，(按位或)。输出这样数列的个数mod 1000000009。

　　　n <= 10，r <= 15000。

解法：先按位分析这道题，若将a1,a2..an转化成二进制形式并对齐如下，则可将题目转化为求每一列最多含有一个1，每一行所对应的数小等于r的矩阵有多少个。

　　　　　　　　　　　　　　　　　　　　　　　

　　　这样的话，下意识地想到用状态压缩的DP来做，但是这样做的时间复杂度为O(10×15000×15000)，不能接受。最后我也没想出更好的方法，只好看了题解。

　　　对于某一行，若要使得它对应的数小于r，只需要在某一列，r的二进制形式为1，它为0；在这一列之前， 该行所有值与r相同；在这之后，该行每一列可以为任意值。

　　　设数组d[i][j]表示从左向右扫描的情况下，从第i位扫到第0位，已经有(n-j)个数小于r的情况下，共有多少个符合题意的数列。具体状态转移方程可以看我得代码，详细的注解见官方题解的代码，http://apps.topcoder.com/wiki/display/tc/SRM+508。

tag:dp, good

// BEGIN CUT HERE
/*
 * Author:  plum rain
 * score :
 */
/*

 */
// END CUT HERE
#line 11 "YetAnotherORProblem2.cpp"
#include <sstream>
#include <stdexcept>
#include <functional>
#include <iomanip>
#include <numeric>
#include <fstream>
#include <cctype>
#include <iostream>
#include <cstdio>
#include <vector>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <cstdlib>
#include <set>
#include <queue>
#include <bitset>
#include <list>
#include <string>
#include <utility>
#include <map>
#include <ctime>
#include <stack>

using namespace std;

#define CLR(x) memset(x, 0, sizeof(x))
#define CLR1(x) memset(x, -1, sizeof(x))
#define PB push_back
#define SZ(v) ((int)(v).size())
#define zero(x) (((x)>0?(x):-(x))<eps)
#define out(x) cout<<#x<<":"<<(x)<<endl
#define tst(a) cout<<#a<<endl
#define CINBEQUICKER std::ios::sync_with_stdio(false)

typedef vector<int> VI;
typedef vector<string> VS;
typedef vector<double> VD;
typedef long long int64;
typedef pair<int, int> pii;

const double eps = 1e-8;
const double PI = atan(1.0)*4;
const int maxint = 2139062143;
const int mod = 1000000009;

int r, n;
int64 d[35][15];

int64 rec (int t, int num)
{
    if (t == -1) return 1;

    int64 &ret = d[t][num];
    int tmp = r & (1 << t);

    if (ret != -1) return ret;

    if (num == n){
        if (tmp)
            return ret = (rec(t-1, 0) + n * rec(t-1, 1)) % mod;
        return ret = rec(t-1, num); 
    }
    if (num == 1){
        if (tmp)
            return ret = (rec(t-1, 0) + rec(t-1, 1) + (n-1) * rec(t-1, 0)) % mod;
        return ret = n * rec(t-1, 1) % mod;
    }
    return ret = (n+1) * rec(t-1, 0) % mod;
}

class YetAnotherORProblem2
{
    public:
        int countSequences(int N, int R){
            r = R; n = N;
            CLR1 (d);
            return (int)((rec(30, n)+mod) % mod);
        }
        
// BEGIN CUT HERE
    public:
    void run_test(int Case) { if ((Case == -1) || (Case == 0)) test_case_0(); if ((Case == -1) || (Case == 1)) test_case_1(); if ((Case == -1) || (Case == 2)) test_case_2(); if ((Case == -1) || (Case == 3)) test_case_3(); }
    private:
    template <typename T> string print_array(const vector<T> &V) { ostringstream os; os << "{ "; for (typename vector<T>::const_iterator iter = V.begin(); iter != V.end(); ++iter) os << '"' << *iter << "","; os << " }"; return os.str(); }
    void verify_case(int Case, const int &Expected, const int &Received) { cerr << "Test Case #" << Case << "..."; if (Expected == Received) cerr << "PASSED" << endl; else { cerr << "FAILED" << endl; cerr << "	Expected: "" << Expected << '"' << endl; cerr << "	Received: "" << Received << '"' << endl; } }
    //void test_case_0() { int Arg0 = 2; int Arg1 = 15000; int Arg2 = 4628299; verify_case(0, Arg2, countSequences(Arg0, Arg1)); }
    void test_case_0() { int Arg0 = 2; int Arg1 = 2; int Arg2 = 7; verify_case(0, Arg2, countSequences(Arg0, Arg1)); }
    void test_case_1() { int Arg0 = 2; int Arg1 = 3; int Arg2 = 9; verify_case(1, Arg2, countSequences(Arg0, Arg1)); }
    void test_case_2() { int Arg0 = 3; int Arg1 = 3; int Arg2 = 16; verify_case(2, Arg2, countSequences(Arg0, Arg1)); }
    void test_case_3() { int Arg0 = 7; int Arg1 = 1023; int Arg2 = 73741815; verify_case(3, Arg2, countSequences(Arg0, Arg1)); }

// END CUT HERE

};

// BEGIN CUT HERE
int main()
{
//    freopen( "a.out" , "w" , stdout );    
    YetAnotherORProblem2 ___test;
    ___test.run_test(-1);
       return 0;
}
// END CUT HERE