Contest2037 - CSU Monthly 2013 Oct (problem B ：Scoop water)

http://acm.csu.edu.cn/OnlineJudge/problem.php?cid=2037&pid=1

【题解】：卡特兰数取模

h(n) = h(n-1)*(4*n-2)/(n+1)

这题我们公式选择错了，不过还是能AC的

因为要取模，应该选 h(n)=h(0)*h(n-1)+h(1)*h(n-2)+...+h(n-1)*h(0) (n>=2)

【code-java】：

 import java.math.BigInteger;
 import java.util.Scanner;

 public class Main {

         public static void main(String[] args) {
             Scanner cin = new Scanner(System.in);
             BigInteger [] arr = new BigInteger[10100];
             BigInteger temp;
             arr[1] = new BigInteger("1");
             for(int i=2;i<=10000;i++){
                 temp = new BigInteger(""+(4*i-2)+"");
                 arr[i] = arr[i-1].multiply(temp).divide(new BigInteger(""+(i+1)+""));
                 //System.out.println(arr[i]);
             }

             while(cin.hasNextInt())
             {
                 BigInteger a;
                 int n,i;
                 n=cin.nextInt();
                 System.out.println(arr[n].mod(new BigInteger("1000000007")));
             }
         }
 }

【code-C++】：

 #include <iostream>
 #include <vector>
 #include <list>
 #include <deque>
 #include <queue>
 #include <iterator>
 #include <stack>
 #include <map>
 #include <set>
 #include <algorithm>
 #include <cctype>
 #include <cstdio>
 #include <cstdlib>
 #include <cstring>
 #include <string>
 #include <cmath>
 using namespace std;

 const int m=;
 typedef long long LL;

 int n;

 void mul(LL &res,int k)
 {
     res=(res*k)%m;
 }

 int ext_gcd(int a,int b,int &x,int &y)
 {
     int ret;
     if(!b)
     {
         x=;y=;return a;
     }
     ret=ext_gcd(b,a%b,y,x);
     y-=x*(a/b);
     return ret;
 }

 void chu(LL &res,int k)
 {
     if(k!=)
     {
         int x,y,temp;
         temp=ext_gcd(k,m,x,y);
         x=(x%m+m)%m;
         res=(res*x)%m;
     }
 }

 LL arr[];

 int main()
 {
      LL res=,l=;
      arr[]=;
     for(int i=;i<=;i++)
     {
         mul(res,*i-);
         chu(res,i+);
         l=res;
         l=l%m;
         arr[i]=l;
     }
     while(scanf("%d",&n)!=EOF)
     {

         printf("%lld\n",arr[n]);
     }
     return ;
 }