【HDU】【The King’s Ups and Downs】

The King’s Ups and Downs

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 366 Accepted Submission(s): 245



Problem Description

The king has guards of all different heights. Rather than line them up in increasing or decreasing height order, he wants to line them up so each guard is either shorter than the guards next to him or taller than the guards next to him (so the heights go up
and down along the line). For example, seven guards of heights 160, 162, 164, 166, 168, 170 and 172 cm. could be arranged as:



or perhaps:



The king wants to know how many guards he needs so he can have a different up and down order at each changing of the guard for rest of his reign. To be able to do this, he needs to know for a given number of guards, n, how many different up and down orders
there are:

For example, if there are four guards: 1, 2, 3,4 can be arrange as:

1324, 2143, 3142, 2314, 3412, 4231, 4132, 2413, 3241, 1423

For this problem, you will write a program that takes as input a positive integer n, the number of guards and returns the number of up and down orders for n guards of differing heights.



Input

The first line of input contains a single integer P, (1 <= P <= 1000), which is the number of data sets that follow. Each data set consists of single line of input containing two integers. The first integer, D is the data set number. The second integer, n (1
<= n <= 20), is the number of guards of differing heights.



Output

For each data set there is one line of output. It contains the data set number (D) followed by a single space, followed by the number of up and down orders for the n guards.



Sample Input

4
1 1
2 3
3 4
4 20

Sample Output

1 1
2 4
3 10
4 740742376475050

Source

Greater New York 2012



Recommend

liuyiding | We have carefully selected several similar problems for you: 5283 5282 5281 5280 5279

题意：

n个不同数字，一大一小一大 | 一小一大一小得排列下去。 求方案的可能总数。

思路：

设F【n】为 n个不同数字的时候满足条件的方案总数。 我们开始枚举最大的那个数的位置。

最大的那个数要不在第一个位置要不在第二个位置。。。最后一个位置。

不妨设最大的数在第j个位置,那么我们可以知道是这样排列的

/ \ / \ / \ * / \ / \

|

位置 j

然后就是F[j-1]/ 2*F[n-j]/2*C[n-1][j-1]

所以嘛,就是这样的。

#include <iostream>
#include <cstring>
#include <cmath>
#include <queue>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <string>
#include <cstdlib>
#include <cstdio>
#include <algorithm>
using namespace std;

typedef long long LL;
const int maxn = 25;
LL C[maxn][maxn];
LL cal()
{
	memset(C,0,sizeof(C));
	C[0][0] = 1;
	for(int i=0;i<22;i++)
	{
		C[i][0] = 1;
	}
	for(int i=1;i<22;i++)
	{
		for(int j=1;j<=i;j++)
		{
			C[i][j] = C[i-1][j] + C[i-1][j-1];
		}
	}
}

int T;
int idx,n;
double sum[maxn];
double dp[maxn][2];
int main()
{
	//freopen("1.txt","r",stdin);
	/*cal();
	int n,m;
	while(cin >> m >> n)
	{
		cout << C[m]
 << endl;
	}
	*/
	cal();
	memset(sum,0,sizeof(sum));
	memset(dp,0,sizeof(dp));
	sum[0] = 2;
	sum[1] = 2;
	dp[1][0] = 1;
	dp[1][1] = 1;
	dp[0][0] = 1;
	dp[0][1] = 1; 

	for(int i=2;i<=20;i++)
	{
    	//sum[i] += dp[1-1][0]*dp[i-1][1]*C[i-1][1-1];

		for(int j=1;j<=i;j++)
		sum[i] += sum[j-1]/2*sum[i-j]/2*C[i-1][j-1];
		dp[i][0] = dp[i][1] = sum[i] / 2;
	}

	scanf("%d",&T);
	while(T --)
	{
		scanf("%d%d",&idx,&n);
		LL num = (long long)sum
;
	//	cout << sum
 << endl;
		if(n == 1)
			num = 1;
		printf("%d %I64d\n",idx,num);
	}

}

------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

#include <iostream>
#include <cstring>
#include <cmath>
#include <queue>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <string>
#include <cstdlib>
#include <cstdio>
#include <algorithm>
using namespace std;

typedef long long LL;
const int maxn = 25;
LL C[maxn][maxn];
LL cal()
{
	memset(C,0,sizeof(C));
	C[0][0] = 1;
	for(int i=0;i<22;i++)
	{
		C[i][0] = 1;
	}
	for(int i=1;i<22;i++)
	{
		for(int j=1;j<=i;j++)
		{
			C[i][j] = C[i-1][j] + C[i-1][j-1];
		}
	}
}

int T;
int idx,n;
LL sum[maxn];
LL dp[maxn][2];
int main()
{
	freopen("1.txt","r",stdin);
	/*cal();
	int n,m;
	while(cin >> m >> n)
	{
		cout << C[m]
 << endl;
	}
	*/
	cal();
	memset(sum,0,sizeof(sum));
	memset(dp,0,sizeof(dp));
	sum[1] = 1;
	dp[1][0] = 1;
	dp[1][1] = 1;
	dp[0][0] = 1;
	dp[0][1] = 1; 

	for(int i=2;i<=20;i++)
	{
		for(int j=1;j<=i;j++)
		sum[i] += dp[j-1][0]*dp[i-j][1]*C[i-1][j-1];
		dp[i][0] = dp[i][1] = sum[i] / 2;
	}

	scanf("%d",&T);
	while(T --)
	{
		scanf("%d%d",&idx,&n);
		printf("%d %I64d\n",idx,sum
);
	}

}