Buy the Ticket

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 5566 Accepted Submission(s): 2326

Problem Description

The "Harry Potter and the Goblet of Fire" will be on show in the next few days. As a crazy fan of Harry Potter, you will go to the cinema and have the first sight, won’t you?

Suppose the cinema only has one ticket-office and the price for per-ticket is 50 dollars. The queue for buying the tickets is consisted of m + n persons (m persons each only has the 50-dollar bill and n persons each only has the 100-dollar bill).

Now the problem for you is to calculate the number of different ways of the queue that the buying process won't be stopped from the first person till the last person.

Note: initially the ticket-office has no money.

The buying process will be stopped on the occasion that the ticket-office has no 50-dollar bill but the first person of the queue only has the 100-dollar bill.

Input

The input file contains several test cases. Each test case is made up of two integer numbers: m and n. It is terminated by m = n = 0. Otherwise, m, n <=100.

Output

For each test case, first print the test number (counting from 1) in one line, then output the number of different ways in another line.

Sample Input

Sample Output

Test #1:

6

Test #2:

18

Test #3:

180

#include<iostream>

#include<cstdio>

#include<cstring>

using namespace std;

#define base 10000

int a[101];

void multiply(int x)

{

a[100]=1;

for(int i=1;i<=x;i++)

for(int j=100,temp=0;j>=0;j--)

{

temp+=a[j]*i;

a[j]=temp%base;

temp=temp/base;

}

}

void chen(int x)

{

int temp=0;

for(int i=100;i>=0;i--)

{

temp+=a[i]*x;

a[i]=temp%base;

temp=temp/base;

}

}

void output()

{

int i=1;

while(!a[i])

i++;

printf("%d",a[i]);//********假如第一位a[1]=34，a[2]=2039，则只能输出343029，而不能输出00342039，所以a[1]需 // 要单独列出来输出

for(i++;i<=100;i++)

printf("%04d",a[i]);//右对齐，左边不足的位数补0，例如假设,a[19]=1234,a[20]=34，a{21]=789则这一段的输出

//就为123400340789；

printf("\n");

}

void chu(int x)

{

int i=1,temp=0;

while(!a[i])

i++;

for(;i<=100;i++)

{

temp=a[i]+temp*base;

a[i]=temp/x;

temp%=x;

}

}

int main()

{

int m,n,cas=0;

while(~scanf("%d %d",&m,&n)&&(m||n))

{

cas++;

printf("Test #%d:\n",cas);

if(n>m) {cout<<"0"<<endl;}

else{

memset(a,0,sizeof(a));

multiply(m+n);

chen(m-n+1);

chu(m+1);

output();

}

}

return 0;

}

/*****受益匪浅的一道题，这道题综合性还是有点强的，首先卡特兰数推出公示后（公式我推了出来），但是因为数字实在太大（毕竟100的阶乘），所以必需对大数进行处理，也就是高精度，很有技巧性的是，可以让数组一个单元存储一万，而不是以前习惯的十（毕竟数太大），最后还有一个非常关键的地方，就是最终的输出（我个人觉得输出才是本题高潮，最有戏剧性的一部分），具体输出可见代码上的注释,最后，悲催的是发现自己很多题目没有把握都是因为不会时间复杂度以及内存存储大小的分析，是时候好好看看这两个知识点了****/

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巴特西

hdu 1133 卡特兰 高精度

Buy the Ticket

最新文章

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hdu 1133 卡特兰高精度