Pills
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 42 Accepted Submission(s): 36
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
Source
The 2011 Rocky Mountain Regional Contest
Recommend
lcy
解题报告：做这道题的时候以为是递归呢！结果推了一下推不出来，今天杭电比赛不知道，来的比较晚，看队友做出来了，就问了一下队友，其实这道题是一道动态规划的题目，每个状态都和前面的状态有关。为了好描述这道题我画了一张表，如下：就是把每一行加起来即为所求。
代码如下：
#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int N = 32;
__int64 dp[N][N];//这儿也得是64位
void DP()
{
int i , j;
memset(dp, 0, sizeof(dp));
dp[0][0] = 1;//初始化
for (i = 1; i <= 30; ++i)//初始化
{
dp[i][1] = 1;
}
for (i = 1; i <= 30; ++i)
{
for (j = 2; j <= i; j ++)
{
dp[i][j] = dp[i - 1][j] + dp[i][j - 1];//状态方程
}
}
}
int main()
{
int n, i;
__int64 ans;
DP();
while (scanf("%d", &n) != EOF &&n)
{
ans = 0;
for (i = 1; i <= n; ++i)
{
ans += dp[n][i];
}
printf("%I64d\n", ans);
}
return 0;
}