SRM 502(2-1000pt)

题意：在0~(n-1)中选择k个数，使得他们的和为n的倍数的选择方案有多少种。(n <= 1000, k <= 47)

解法：裸dp。d[i][j][k’]表示在前i个数中(0~i-1)，选择k‘个数使得其和mod n的余数为j的选择方案的种数。时间、空间复杂度均为O(n^2*k)，时间复杂度能接受，空间不能，用滚动数组优化即可。

tag:dp, 水题

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

  */
 // END CUT HERE
 #line 11 "TheCowDivTwo.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 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;

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

 int d[][][];

 class TheCowDivTwo
 {
     public:
         int find(int n, int K){
             CLR (d);
             d[][][] = ;
             int cur = ;
             for (int i = ; i <= n; ++ i){
                 for (int j = ; j < n; ++ j){
                     int tmp = j-(i-)< ? n+j-(i-) : j-(i-);
                     for (int k = ; k <= K; ++ k){
                         d[cur][j][k] = d[cur^][j][k];
                         if (k){
                             d[cur][j][k] = (d[cur][j][k] + d[cur^][tmp][k-]) % mod;
                         }
                     }
                 }
                 cur ^= ;
             }

             return d[cur^][][K];
         }

 // 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(); if ((Case == -) || (Case == )) test_case_4(); }
     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 = ; int Arg1 = ; int Arg2 = ; verify_case(, Arg2, find(Arg0, Arg1)); }
     void test_case_1() { int Arg0 = ; int Arg1 = ; int Arg2 = ; verify_case(, Arg2, find(Arg0, Arg1)); }
     void test_case_2() { int Arg0 = ; int Arg1 = ; int Arg2 = ; verify_case(, Arg2, find(Arg0, Arg1)); }
     void test_case_3() { int Arg0 = ; int Arg1 = ; int Arg2 = ; verify_case(, Arg2, find(Arg0, Arg1)); }
     void test_case_4() { int Arg0 = ; int Arg1 = ; int Arg2 = ; verify_case(, Arg2, find(Arg0, Arg1)); }

 // END CUT HERE

 };

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