题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4489点击打开链接

The King’s Ups and Downs

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 684 Accepted Submission(s): 492

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

解释起来有点复杂

首先这是一个像波一样的我们将高的称做波峰矮的称做波谷

当新增加的一人插在队伍两边的时候 a[i]+=a[i-1]

当新增的人插在中间的时候我们就对其中的一段进行排列组合

同时注意选取的这两段的两端必须同为波峰或者波谷而这种情况出现为当前这段总数的1/2

因此就可以知道当前的a[i] 除了放在两边意外还有内部所有排列组合可能出现的情况的递推而这些就是计算新增加元素在内部的个数

注意排列组合其中一段为1时特判

#include <iostream>
#include <queue>
#include <stdio.h>
#include <stdlib.h>
#include <stack>
#include <limits>
#include <string>
#include <string.h>
#include <vector>
#include <set>
#include <map>
#include <algorithm>
#include <math.h>
using namespace std;
long long int jc(int x)
{long long int sum=1;for(int i=2;i<=x;i++)sum*=i;return sum;
}
long long int c(int x,int y)
{return jc(x)/jc(y)/jc(x-y);
}
int main()
{long long int a[60];memset(a,0,sizeof(a));a[1]=1;a[2]=2;a[3]=4;for(int i=4;i<=20;i++){a[i]+=a[i-1];for(int j=1;j<i-1;j++){if(j==1||i-1-j==1)a[i]+=(a[j]*a[i-j-1]>>1)*c(i-1,j);elsea[i]+=(a[j]>>1)*(a[i-j-1]>>1)*c(i-1,j);}}int n;scanf("%d",&n);while(n--){int mid,m;scanf("%d%d",&m,&mid);printf("%d %lld\n",m,a[mid]);}}

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