CSU 1320 scoop water (catlan number)

1320: scoop watertime limit: 2 sec memory limit: 128 MB
Submit: 494 solved: 129
[Submit] [Status] [web board] Description

I just bought two water polo a and B today, and the capacity is 1 litre. He plans to use this water polo to play games.

First, the supervisor has prepared a water tank with an infinite capacity. At the beginning, the water tank is empty. Then, the water tank A is used to add water to the water tank, and the water tank B is used to scoop out the water tank, when using water tank B to scoop out the water, there must be at least 1 litre of water in the water tank. In this way, the water tank will remain empty after N times of water diversion A and N times of water diversion B.

Input

Input Multiple examples until the end of the file.

Each example contains only one n (0 <n <= 10000), indicating that you must use n times of a watermark and N times of B watermark.

Output

For each example, please output a number, indicating the total number of correct draft methods so that there will always be enough water in the water tank when B is used in the draft process.

(Because the number is large, the output answer model is 1000000007)

Sample Input

12

Sample output

12

Hint

 

Csu_cx

 

Source

CSU monthly 2013 Oct.

 

Catlan number's application on the order of output stacks

 1 #include<cstdio> 2 #include<iostream> 3 #include<cstring> 4 #include<stdlib.h> 5 #include<algorithm> 6 using namespace std; 7 const int mod=1000000007; 8 long long f[10005]; 9 int n;10 int main()11 {12     //freopen("in.txt","r",stdin);;13     f[0]=1;14     for(int i=1;i<=10000;i++)15         for(int j=0;j<i;j++)16             f[i]=((f[j]*f[i-1-j])%mod+f[i])%mod;17             18     while(scanf("%d",&n)!=EOF)19     {20         printf("%lld\n",f[n]);21     }22     return 0;23 }

View code

 

CSU 1320 scoop water (catlan number)