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

  1 // BEGIN CUT HERE
  2 /*
  3  * Author:  plum rain
  4  * score :
  5  */
  6 /*
  7 
  8  */
  9 // END CUT HERE
 10 #line 11 "YetAnotherORProblem2.cpp"
 11 #include <sstream>
 12 #include <stdexcept>
 13 #include <functional>
 14 #include <iomanip>
 15 #include <numeric>
 16 #include <fstream>
 17 #include <cctype>
 18 #include <iostream>
 19 #include <cstdio>
 20 #include <vector>
 21 #include <cstring>
 22 #include <cmath>
 23 #include <algorithm>
 24 #include <cstdlib>
 25 #include <set>
 26 #include <queue>
 27 #include <bitset>
 28 #include <list>
 29 #include <string>
 30 #include <utility>
 31 #include <map>
 32 #include <ctime>
 33 #include <stack>
 34 
 35 using namespace std;
 36 
 37 #define CLR(x) memset(x, 0, sizeof(x))
 38 #define CLR1(x) memset(x, -1, sizeof(x))
 39 #define PB push_back
 40 #define SZ(v) ((int)(v).size())
 41 #define zero(x) (((x)>0?(x):-(x))<eps)
 42 #define out(x) cout<<#x<<":"<<(x)<<endl
 43 #define tst(a) cout<<#a<<endl
 44 #define CINBEQUICKER std::ios::sync_with_stdio(false)
 45 
 46 typedef vector<int> VI;
 47 typedef vector<string> VS;
 48 typedef vector<double> VD;
 49 typedef long long int64;
 50 typedef pair<int, int> pii;
 51 
 52 const double eps = 1e-8;
 53 const double PI = atan(1.0)*4;
 54 const int maxint = 2139062143;
 55 const int mod = 1000000009;
 56 
 57 int r, n;
 58 int64 d[35][15];
 59 
 60 int64 rec (int t, int num)
 61 {
 62     if (t == -1) return 1;
 63 
 64     int64 &ret = d[t][num];
 65     int tmp = r & (1 << t);
 66 
 67     if (ret != -1) return ret;
 68 
 69     if (num == n){
 70         if (tmp)
 71             return ret = (rec(t-1, 0) + n * rec(t-1, 1)) % mod;
 72         return ret = rec(t-1, num); 
 73     }
 74     if (num == 1){
 75         if (tmp)
 76             return ret = (rec(t-1, 0) + rec(t-1, 1) + (n-1) * rec(t-1, 0)) % mod;
 77         return ret = n * rec(t-1, 1) % mod;
 78     }
 79     return ret = (n+1) * rec(t-1, 0) % mod;
 80 }
 81 
 82 class YetAnotherORProblem2
 83 {
 84     public:
 85         int countSequences(int N, int R){
 86             r = R; n = N;
 87             CLR1 (d);
 88             return (int)((rec(30, n)+mod) % mod);
 89         }
 90         
 91 // BEGIN CUT HERE
 92     public:
 93     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(); }
 94     private:
 95     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(); }
 96     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; } }
 97     //void test_case_0() { int Arg0 = 2; int Arg1 = 15000; int Arg2 = 4628299; verify_case(0, Arg2, countSequences(Arg0, Arg1)); }
 98     void test_case_0() { int Arg0 = 2; int Arg1 = 2; int Arg2 = 7; verify_case(0, Arg2, countSequences(Arg0, Arg1)); }
 99     void test_case_1() { int Arg0 = 2; int Arg1 = 3; int Arg2 = 9; verify_case(1, Arg2, countSequences(Arg0, Arg1)); }
100     void test_case_2() { int Arg0 = 3; int Arg1 = 3; int Arg2 = 16; verify_case(2, Arg2, countSequences(Arg0, Arg1)); }
101     void test_case_3() { int Arg0 = 7; int Arg1 = 1023; int Arg2 = 73741815; verify_case(3, Arg2, countSequences(Arg0, Arg1)); }
102 
103 // END CUT HERE
104 
105 };
106 
107 // BEGIN CUT HERE
108 int main()
109 {
110 //    freopen( "a.out" , "w" , stdout );    
111     YetAnotherORProblem2 ___test;
112     ___test.run_test(-1);
113        return 0;
114 }
115 // END CUT HERE
View Code

 

 

 

posted @ 2013-12-13 20:18  Plumrain  阅读(180)  评论(0)  编辑  收藏  举报