• HDU 1023 Train Problem II


    Train Problem II

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
    Total Submission(s): 2830    Accepted Submission(s): 1562

    Problem Description
    As we all know the Train Problem I, the boss of the Ignatius Train Station want to know if all the trains come in strict-increasing order, how many orders that all the trains can get out of the railway.
     
    Input
    The input contains several test cases. Each test cases consists of a number N(1<=N<=100). The input is terminated by the end of file.
     
    Output
    For each test case, you should output how many ways that all the trains can get out of the railway.
     
    Sample Input
    1
    2
    3
    10
     
    Sample Output

    1

    2

    5

    16796

    卡特兰数  大数乘除

    Java版本

    import java.util.*;
    import java.io.*;
    import java.math.*;
    public class Main
    {
        static Scanner cin=new Scanner(System.in);
        static PrintWriter cout=new PrintWriter(System.out,true);
        public static void main(String[] args)
        {
            BigInteger a[]=new BigInteger[110];
            BigInteger b=BigInteger.valueOf(4);
            BigInteger d=BigInteger.valueOf(2);
            a[1]=BigInteger.ONE;
            for(int i=2;i<=100;i++)
            {
                BigInteger c=BigInteger.valueOf(i);
                BigInteger e=BigInteger.valueOf(i+1);
                a[i]=a[i-1].multiply(b.multiply(c).subtract(d)).divide(e);
            }
            while(cin.hasNext())
            {
                int n=cin.nextInt();
                System.out.println(a[n].toString());
            }
        }
    }

    c++版本

    #include <iostream> 
    #include <algorithm> 
    #include <cstring> 
    #include <cstdio>
    #include <vector> 
    #include <queue> 
    #include <cstdlib> 
    #include <iomanip>
    #include <cmath> 
    #include <ctime> 
    #include <map> 
    #include <set> 
    using namespace std; 
    #define lowbit(x) (x&(-x)) 
    #define max(x,y) (x>y?x:y) 
    #define min(x,y) (x<y?x:y) 
    #define MAX 100000000000000000 
    #define MOD 1000000007
    #define pi acos(-1.0) 
    #define ei exp(1) 
    #define PI 3.141592653589793238462
    #define ios() ios::sync_with_stdio(false)
    #define INF 1044266558
    #define mem(a) (memset(a,0,sizeof(a)))
    typedef long long ll;
    //f[n]=2*(2*n-1)*f[n-1]/(i+1);
    int a[101][101],n;
    void multiply(int *a,int k,int b)
    {
        int ans=0;
        for(int i=k-1;i>=0;i--)
        {
            ans+=a[i]*b;
            a[i]=ans%10000;
            ans/=10000;
        }
    }
    void divide(int *a,int k,int b)
    {
        int ans=0;
        for(int i=0;i<k;i++)
        {
            ans*=10000;
            ans+=a[i];
            a[i]=ans/b;
            ans%=b;
        }
    }
    void init()
    {
        memset(a,0,sizeof(a));
        a[1][99]=1;
        for(int i=2;i<=100;i++)
        {
            memcpy(a[i],a[i-1],sizeof(a[i-1]));
            multiply(a[i],100,4*i-2);
            divide(a[i],100,i+1);
        }
    }
    int main()
    {
        init();
        int i;
        while(scanf("%d",&n)!=EOF)
        {
            for(i=0;i<100 && !a[n][i];i++);
            printf("%d",a[n][i++]);
            for(;i<100;i++)
                printf("%04d",a[n][i]);
            printf("
    ");
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/shinianhuanniyijuhaojiubujian/p/7299441.html
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