SRM 609 D2L3: VocaloidsAndSongs，dp

题目：http://community.topcoder.com/stat?c=problem_statement&pm=12989&rd=15842

不算难，注意 base case是 cur == S，cur表示当前已经唱了的歌曲数。

代码:

#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)

/*************** Program Begin **********************/
int dp[51][51][51][51];
int d1[] = {1, 0, 1, 0, 1, 0, 1};
int d2[] = {0, 1, 1, 0, 0, 1, 1};
int d3[] = {0, 0, 0, 1, 1, 1, 1};
const int MOD = 1e9 + 7;
class VocaloidsAndSongs {
public:
	int S, gumi, ia, mayu;
	int rec(int cur, int v1, int v2, int v3)
	{
		if (cur == S) { 
			if (v1 == gumi && v2 == ia && v3 == mayu) {
				return 1;
			} else {
				return 0;
			}
		}
		int & res = dp[cur][v1][v2][v3];
		if (res != -1) {
			return res;
		}
		res = 0;
		for (int i = 0; i < 7; i++) {
			int t1 = v1 + d1[i];
			int t2 = v2 + d2[i];
			int t3 = v3 + d3[i];
			if (t1 > gumi || t2 > ia || t3 > mayu) {
				continue;
			} else {
				res += rec(cur + 1, t1, t2, t3);
				res %= MOD;
			}
		}
		return res;
	}
	int count(int S, int gumi, int ia, int mayu) {
		this->S = S;
		this->gumi = gumi;
		this->ia = ia;
		this->mayu = mayu;
		memset(dp, -1, sizeof(dp));
		return rec(0, 0, 0, 0);
	}

};

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

SRM 609 D2L3: VocaloidsAndSongs，dp

原文：http://blog.csdn.net/xzz_hust/article/details/19365999