D - Permutation Counting

http://acm.hdu.edu.cn/showproblem.php?pid=6880

根据长度为n的排列a，构造长度n-1的序列b

思路：DP

官方题解：

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<bitset>
#include<cassert>
#include<cctype>
#include<cmath>
#include<cstdlib>
#include<ctime>
#include<deque>
#include<iomanip>
#include<list>
#include<map>
#include<queue>
#include<set>
#include<stack>
#include<vector>
#include <vector>
#include <iterator>
#include <utility>
#include <sstream>
#include <limits>
#include <numeric>
#include <functional>
using namespace std;
#define gc getchar()
#define mem(a) memset(a,0,sizeof(a))
#define debug(x) cout<<"debug:"<<#x<<" = "<<x<<endl;

#define ios ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);

typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
typedef pair<int,int> pii;
typedef char ch;
typedef double db;

const double PI=acos(-1.0);
const double eps=1e-6;
const int inf=0x3f3f3f3f;
//const int maxn=1e5+10;
const int maxn = 5010;
//const int maxm=100+10;
const int N=1e6+10;
const int mod=1e9+7;

int A[maxn] = {0};
int P[maxn][maxn] = {0};
int main()
{
	int n = 0 , ans = 0;
	int T = 0;
	cin >> T;

	while(T--)
	{
		cin >> n;
		for(int i = 0;i < n;i++)
		{
			if(i < n-1)
			{
				cin >> A[i];
			}
			for(int j = 0;j < n;j++)
			{
			 	P[i][j] = 0;
			}
		}
		P[1][0] =  A[0];
		P[1][1] = !A[0];
		for(int i = 1;i < n-1;i++)
		{
			if(A[i])
			{
				for(int j = 0;j < i;j++)
				{
					P[i+1][i-j] = (P[i+1][i-j+1] + P[i][i-j]) % mod;
				}
				P[i+1][i+1] = 0;
			}
			else
			{
			 	P[i+1][1] = 0;
				for(int j = 1;j < i+1;j++)
				{
					P[i+1][j] = (P[i][j-1] + P[i+1][j-1]) % mod;
				}
			}
		}
		for(int i = 0;i < n;i++)
		{
		 	ans = (ans + P[n][i]) % mod;
		}
		cout << ans << endl;
	}
    return 0;
}