Problem E

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 0    Accepted Submission(s): 0

Problem Description Aunt Lizzie takes half a pill of a certain medicine every day. She starts with a bottle that contains N pills.

On the first day, she removes a random pill, breaks it in two halves, takes one half and puts the other half back into the bottle.

On subsequent days, she removes a random piece (which can be either a whole pill or half a pill) from the bottle. If it is half a pill, she takes it. If it is a whole pill, she takes one half and puts the other half back into the bottle.

In how many ways can she empty the bottle? We represent the sequence of pills removed from the bottle in the course of 2N days as a string, where the i-th character is W if a whole pill was chosen on the i-th day, and H if a half pill was chosen (0 <= i < 2N). How many different valid strings are there that empty the bottle?  

Input The input will contain data for at most 1000 problem instances. For each problem instance there will be one line of input: a positive integer N <= 30, the number of pills initially in the bottle. End of input will be indicated by 0.  

Output For each problem instance, the output will be a single number, displayed at the beginning of a new line. It will be the number of different ways the bottle can be emptied.  

Sample Input 6 1 4 2 3 30 0  

Sample Output 132 1 14 2 5 3814986502092304     相当于求N个W，和N个H的排列数，而且要求前面任意个中必需H的个数不大于W的个数~~~~ 不知道哪些大神是怎么样做出来的，好快的速度就做出来了~~~~   我想了好久都没有想到DP的方法， 最后想出来了，感觉方法比较笨~~~~   转化成了图，W表示往下走，H表是往右走， 则必须在左下角中。 N个W和H的时候就是到N-1行的路径数 如N=1时，为dp[0,0] N=2时为dp[1,0]+dp[1,1] N=3时为dp[2,0]+dp[2,1]+dp[2,2]     应该会有更好的方法的~~~~~ 我的思路有点蹉了~~~~但幸好做出来了！   代码如下：

#include<stdio.h>
long long dp[31][31];
void init()
{
    dp[0][0]=1;
    for(int i=1;i<31;i++)
    {
        dp[i][0]=1;
        for(int j=1;j<i;j++)
          dp[i][j]=dp[i-1][j]+dp[i][j-1];
        dp[i][i]=dp[i][i-1];
    }   
} 
int main()
{
    int N;
    init();
    while(scanf("%d",&N),N)
    {
        printf("%I64d\n",dp[N][N]);
    }    
    return 0;
}       

要求的其实就是从（0，0）到(N,N)的路径数，只能往下或者往右走，而且不能走到右上部分去。