【组合数学】TopCoder SRM555 MuddyRoad2

非常精彩的一道组合数学题目，是作为division2的压轴题，包括了斐波那契数列、动态规划、组合数求解等多项知识。

题目在http://community.topcoder.com/stat?c=problem_statement&pm=12189&rd=15177

#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include<cstring>

#define mod 555555555

using namespace std;

class MuddyRoad2 {
public:
	int theCount(int, int);
};


int MuddyRoad2::theCount(int N, int muddyCount) {
		int combination[560][560];
		int result;
		for(int i=0;i<560;i++)
			for(int j=0;j<560;j++)
				combination[i][j]=0;
		//memset(combination,0,sizeof(combination));
		//prepare the combination number
		for(int i=1;i<=N-2;i++){
			combination[i][0]=1;
			for(int j=1;j<=i;j++){
				if(i==j){
					combination[i][j]=1;
					continue;
				}
				combination[i][j]=(combination[i-1][j]+combination[i-1][j-1])%mod;
			}
		}

		int f[560][560][3];
		for(int i=0;i<560;i++)
			for(int j=0;j<560;j++)
				for(int k=0;k<3;k++)
					f[i][j][k]=0;
		for(int i=1;i<=N;i++){
			f[i][1][i%3]=1;
			for(int j=2;j<=i;j++){
				for(int k=1;k<=2;k++){
					f[i][j][k]=(f[i-k][j-1][1]+f[i-k][j-1][2])%mod;
					if(i>=3){
						f[i][j][k]=(f[i][j][k]+f[i-3][j][k])%mod;
					}
				}//end internal for loop
			}//end external for loop
		}
		
		result=combination[N-2][muddyCount]-f[N-muddyCount][muddyCount+1][1]-f[N-muddyCount][muddyCount+1][2];
		//in case that result may be a negative number
		result=(result+mod*2)%mod;
		return result;
}//end theCount


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