题目：http://community.topcoder.com/stat?c=problem_statement&pm=12891&rd=15713

参考：http://apps.topcoder.com/wiki/display/tc/SRM+601

先尝试简单dp方法，时间复杂度O( max(N, M)^3 )，此方法只能求解 max(N, M)不超过200的情况，如下：

#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <iostream>
#include <sstream>
#include <iomanip>

#include <bitset>
#include <string>
#include <vector>
#include <stack>
#include <deque>
#include <queue>
#include <set>
#include <map>

#include <cstdio>
#include <cstdlib>
#include <cctype>
#include <cmath>
#include <cstring>
#include <ctime>
#include <climits>


using namespace std;



#define CHECKTIME() printf("%.2lf\n", (double)clock() / CLOCKS_PER_SEC)
typedef pair<int, int> pii;
typedef long long llong;
typedef pair<llong, llong> pll;
#define mkp make_pair

/*************** Program Begin **********************/
const int MOD = 1e9 + 7;
int dp[200][200][200];

class WinterAndSnowmen {
public:
	int N, M;
	int rec(int t, int x, int y)
	{
		if (t == 0) {
			if (x < y) {
				return 1;
			} else {
				return 0;
			}
		}
		int & res = dp[t][x][y];
		if (res != -1) {
			return res;
		}
		res = 0;
		if (t <= N) {
			res += rec(t - 1, x ^ t, y);
			res %= MOD;
		}
		if (t <= M) {
			res += rec(t - 1, x, y ^ t);
			res %= MOD;
		}
		res += rec(t - 1, x, y);
		res %= MOD;
		return res;
	}
	int getNumber(int N, int M) {
		this->N = N;
		this->M = M;
		memset(dp, -1, sizeof(dp));
		return rec(max(N, M), 0, 0);
	}

};

/************** Program End ************************/

按 Editorials 中给的方法进行优化，得到O( max(N, M) ^ 2 )，最后的optimizations一定也要做，要不然还是会超时，如下：

#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <iostream>
#include <sstream>
#include <iomanip>

#include <bitset>
#include <string>
#include <vector>
#include <stack>
#include <deque>
#include <queue>
#include <set>
#include <map>

#include <cstdio>
#include <cstdlib>
#include <cctype>
#include <cmath>
#include <cstring>
#include <ctime>
#include <climits>


using namespace std;



#define CHECKTIME() printf("%.2lf\n", (double)clock() / CLOCKS_PER_SEC)
typedef pair<int, int> pii;
typedef long long llong;
typedef pair<llong, llong> pll;
#define mkp make_pair

/*************** Program Begin **********************/
const int MOD = 1e9 + 7;
const int MAX_BITS = 11;
const int MAX_N = 2000;
int dp[MAX_N + 1][1 << MAX_BITS][2];

class WinterAndSnowmen {
public:
	int p, N, M;
	inline int getBit(int z, int p) { return ( (z >> p) & 1 ); }

	int rec(int t, int z, int b)
	{
		if (0 == t) {
			if ( 0 == b && 1 == z ) {
				return 1;
			} else {
				return 0;
			}
		}
		int & res = dp[t][z][b];
		if (res != -1) {
			return res;
		}
		res = 0;
		if (t <= N) {
			res += rec(t - 1, z ^ (t >> p), b ^ getBit(t, p));
			res %= MOD;
		}
		if (t <= M) {
			res += rec(t - 1, z ^ (t >> p), b);
			res %= MOD;
		}
		res += rec(t - 1, z, b);
		res %= MOD;
		return res;
	}
	int getNumber(int N, int M) {
		this->N = N;
		this->M = M;
		int res = 0;
		for (p = 0; p < 11; p++) {
			memset(dp, -1, sizeof(dp));
			res += rec( max(N, M), 0, 0 );
			res %= MOD;
		}
		return res;
	}

};

/************** Program End ************************/

SRM 601 D1 L2：WinterAndSnowmen， dp