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-;
 const double PI = atan(1.0)*;
 const int maxint = ;
 const int mod = ;

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

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

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

     if (ret != -) return ret;

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

 class YetAnotherORProblem2
 {
     public:
         int countSequences(int N, int R){
             r = R; n = N;
             CLR1 (d);
             return (int)((rec(, n)+mod) % mod);
         }

 // BEGIN CUT HERE
     public:
     void run_test(int Case) { if ((Case == -) || (Case == )) test_case_0(); if ((Case == -) || (Case == )) test_case_1(); if ((Case == -) || (Case == )) test_case_2(); if ((Case == -) || (Case == )) 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 << "\tExpected: \"" << Expected << '\"' << endl; cerr << "\tReceived: \"" << 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 = ; int Arg1 = ; int Arg2 = ; verify_case(, Arg2, countSequences(Arg0, Arg1)); }
     void test_case_1() { int Arg0 = ; int Arg1 = ; int Arg2 = ; verify_case(, Arg2, countSequences(Arg0, Arg1)); }
     void test_case_2() { int Arg0 = ; int Arg1 = ; int Arg2 = ; verify_case(, Arg2, countSequences(Arg0, Arg1)); }
     void test_case_3() { int Arg0 = ; int Arg1 = ; int Arg2 = ; verify_case(, Arg2, countSequences(Arg0, Arg1)); }

 // END CUT HERE

 };

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