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-8;
const double PI = atan(1.0)*4;
const int maxint = 2139062143;
const int mod = 1000000007;

int d[2][1005][50];

class TheCowDivTwo
{
    public:
        int find(int n, int K){
            CLR (d);
            d[0][0][0] = 1;
            int cur = 1;
            for (int i = 1; i <= n; ++ i){
                for (int j = 0; j < n; ++ j){
                    int tmp = j-(i-1)<0 ? n+j-(i-1) : j-(i-1);
                    for (int k = 0; k <= K; ++ k){
                        d[cur][j][k] = d[cur^1][j][k];
                        if (k){
                            d[cur][j][k] = (d[cur][j][k] + d[cur^1][tmp][k-1]) % mod;
                        }
                    }
                }
                cur ^= 1;
            }

            return d[cur^1][0][K];
        }
        
// 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(); if ((Case == -1) || (Case == 4)) 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 << "	Expected: "" << Expected << '"' << endl; cerr << "	Received: "" << Received << '"' << endl; } }
    void test_case_0() { int Arg0 = 7; int Arg1 = 4; int Arg2 = 5; verify_case(0, Arg2, find(Arg0, Arg1)); }
    void test_case_1() { int Arg0 = 1; int Arg1 = 1; int Arg2 = 1; verify_case(1, Arg2, find(Arg0, Arg1)); }
    void test_case_2() { int Arg0 = 58; int Arg1 = 4; int Arg2 = 7322; verify_case(2, Arg2, find(Arg0, Arg1)); }
    void test_case_3() { int Arg0 = 502; int Arg1 = 7; int Arg2 = 704466492; verify_case(3, Arg2, find(Arg0, Arg1)); }
    void test_case_4() { int Arg0 = 1000; int Arg1 = 47; int Arg2 = 219736903; verify_case(4, Arg2, find(Arg0, Arg1)); }

// END CUT HERE

};

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