Contest2037

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=1000000007;
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=1;y=0;return a;
    }
    ret=ext_gcd(b,a%b,y,x);
    y-=x*(a/b);
    return ret;
}

void chu(LL &res,int k)
{
    if(k!=1)
    {
        int x,y,temp;
        temp=ext_gcd(k,m,x,y);
        x=(x%m+m)%m;
        res=(res*x)%m;
    }
}

LL arr[11111];

int main()
{
     LL res=1,l=1;
     arr[1]=1;
    for(int i=2;i<=10000;i++)
    {
        mul(res,4*i-2);
        chu(res,i+1);
        l=res;
        l=l%m;
        arr[i]=l;
    }
    while(scanf("%d",&n)!=EOF)
    {

        printf("%lld
",arr[n]);
    }
    return 0;
}