import java.math.BigInteger;
import java.util.*;
public class Main {
public static void main(String []args)
{
Scanner cin=new Scanner(System.in);
int n,m,i;
int t1=0;
while(cin.hasNextBigInteger())
{
t1++;
m=cin.nextInt();
n=cin.nextInt();
if(m==0&&n==0) break;
System.out.println("Test #"+t1+":");
if(n>m)
{
System.out.println("0");
}
else
{
BigInteger sum=new BigInteger("1");
for(i=m+n; i>=1; i--)
{
BigInteger c1=new BigInteger(((Integer)i).toString());
sum=sum.multiply(c1);
}
int c=(m-n+1);
BigInteger m2=new BigInteger(((Integer)(m+1)).toString());
BigInteger m1=new BigInteger(((Integer)c).toString());
System.out.println(sum.multiply(m1).divide(m2));
}
}
} }

hdu Buy the Ticket的更多相关文章

  1. 【HDU 1133】 Buy the Ticket (卡特兰数)

    Buy the Ticket Problem Description The "Harry Potter and the Goblet of Fire" will be on sh ...

  2. hdu 1133 Buy the Ticket(Catalan)

    Buy the Ticket Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) T ...

  3. hdu 1133 Buy the Ticket (大数+递推)

    Buy the Ticket Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)To ...

  4. HDU——1133 Buy the Ticket

    Buy the Ticket Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) T ...

  5. HDU 1133 Buy the Ticket (数学、大数阶乘)

    Buy the Ticket Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)To ...

  6. HDUOJ---1133(卡特兰数扩展)Buy the Ticket

    Buy the Ticket Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)To ...

  7. Buy the Ticket{HDU1133}

    Buy the TicketTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total ...

  8. 【高精度练习+卡特兰数】【Uva1133】Buy the Ticket

    Buy the Ticket Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) T ...

  9. Buy the Ticket(卡特兰数+递推高精度)

    Buy the Ticket Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Tota ...

随机推荐

  1. Codeforces 494D Upgrading Array

    http://codeforces.com/contest/494/problem/D 题意:给一个数组,和一个坏质数集合,可以无数次地让1到i这些所有数字除以他们的gcd,然后要求Σf(a[i])的 ...

  2. LeetCode_Word Ladder

    Given two words (start and end), and a dictionary, find the length of shortest transformation sequen ...

  3. ftp和http转参数的使用(转)

    浏览器因特网资源:URL是浏览器寻找信息时所需的资源位置,通过URL,应用程序才能找到并使用共享因特网上大量的数据资源. 大部分URL都遵循一种标准的格式: ①HTTP协议(http://或者http ...

  4. 使用git pull时,项目没有更新?

    进入项目目录后,执行 git pull 命令,没有将项目更新,并提示下图: 提示:there is no tracking information for the current branch. 意思 ...

  5. July 【补题】

    A(zoj 3596) bfs,记忆搜都可以, 按余数来记录状态. B(zoj 3599) 博弈,跳过 C(zoj 3592) 简单dp,题意不好懂 D(zoj 3602) 子树哈希, 对根的左右儿子 ...

  6. PHP商城购物车类

    <?php /* 购物车类 */ // session_start(); class Cart { //定义一个数组来保存购物车商品 private $iteams; private stati ...

  7. How to check for and disable Java in OS X

    Java used to be deeply embedded in OS X, but in recent versions of the OS it's an optional install. ...

  8. Android 环境配置:git开启多颜色模式

    git config --global color.status autogit config --global color.diff autogit config --global color.br ...

  9. 【网络协议】TCP连接的建立和释放

    转载请注明出处:http://blog.csdn.net/ns_code/article/details/29382883 TCP首部格式 先看TCP报文段的格式,例如以下; TCP报文段首部的前20 ...

  10. [LeetCode] 034. Search for a Range (Medium) (C++/Java)

    索引:[LeetCode] Leetcode 题解索引 (C++/Java/Python/Sql) Github: https://github.com/illuz/leetcode 035. Sea ...