HDU 4489The King’s Ups and Downs(dp递推)

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

41 12 33 44 20

 

Sample Output

    
    

     
     1 1
2 4
3 10
4 740742376475050
    
    

 

      解析：

 

代码：

#include <iostream>
#include <cstring>
#include<cstdio>
using namespace std;
__int64 dp[25][2],c[25][25],ans;

int main()
{
    int i,j,p,t,k;
    for(i=1;i<21;i++)
    {
        c[i][0]=c[i][i]=1;
         for(j=1;j<i;j++)
         {
            c[i][j]=c[i-1][j]+c[i-1][j-1];
//            cout<<"c["<<i<<"]["<<j<<"]"<<c[i][j]<<endl;
         }
    }
    dp[0][0]=1;dp[0][1]=1;
    dp[1][0]=1;dp[1][1]=1;
    for(i=2;i<21;i++)
    {
        ans = 0;
        for(j=0;j<i;j++)
            ans += (dp[j][0] * dp[i-j-1][1] * c[i-1][j]);
        dp[i][0]=dp[i][1]=ans/2;
 //        cout<<"dp["<<i<<"][1]"<<dp[i][1]*2<<endl;
    }
    scanf("%d",&p);
    while(p--)
    {
        scanf("%d %d",&t,&k);
        if(k==1)
            printf("%d 1\n",t);
        else
            printf("%d %I64d\n",t,dp[k][0]<<1);
    }
    return 0;
}
/*
打表
#include <iostream>
#include <cstring>
#include<cstdio>
using namespace std;
__int64 ans[21]={
    0,1,2,4,10,32,122,544,2770,15872,101042,707584,5405530,44736512,398721962,
    3807514624ll,38783024290ll,419730685952ll,4809759350882ll,58177770225664ll,
    740742376475050ll};

int main()
{
    int p,t,k;
    scanf("%d",&p);
    while(p--)
    {
        scanf("%d %d",&t,&k);
        printf("%d %I64d\n",t,ans[k]);
    }
    return 0;
}
*/