Train Problem II
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7198 Accepted Submission(s): 3885
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
Hint
The result will be very large, so you may not process it by 32-bit integers.
#include<stdio.h>
#include<math.h>
#include<string.h>
#define base 10000//四位数字的处理
#define max 120
using namespace std;
int catalan[max][max];
void big_mul(int *a,int n)
{
int temp=0;
for(int i=0;i<max;i++)
{
temp+=a[i]*n;
a[i]=temp%base;
temp/=base;
}
}
void big_div(int *a,int n)
{
int temp=0;
for(int i=max-1;i>=0;i--)
{
temp=temP*base+a[i];
a[i]=temp/n;
temp%=n;
}
}
void set()
{
for(int i=0;i<max;i++)
memset(catalan[i],sizeof(int)*max);
catalan[0][0]=1;
catalan[1][0]=1;
for(int i=2;i<max;i++)
{
memcpy(catalan[i],catalan[i-1],sizeof(int)*max);
big_mul(catalan[i],4*i-2);
big_div(catalan[i],i+1);
}
}
void output(int *a,int n)
{
int i=n-1;
while(a[i]==0) i--;
printf("%d",a[i--]);
while(i>=0) printf("%04d",a[i--]);
printf("\n");
}
int main()
{
set();
int t;
while(~scanf("%d",&t))
{
output(catalan[t],max);
}
/*for(int i=0;i<max;i++)
output(catalan[i],max);
return 0;*/
}
/////////////////////////////下为个人模板0.0
#include<stdio.h>
#include<math.h>
#include<string.h>
#define base 10000
#define max 120
using namespace std;
int cat[max][max];
////此处乘除法按照人正常思路去做0.0
void big_mul(int *a,int n)
{
int temp=0;//进位系统
for(int i=0;i<max;i++)
{
temp+=a[i]*n;
a[i]=temp%base;
temp/=base;
}
}
void big_div(int *a,int n)
{
int temp=0;
for(int i=max-1;i>=0;i--)
{
temp=temP*base+a[i];
a[i]=temp/n;
temp%=n;
}
}
void set()
{
for(int i=0;i<max;i++)
memset(cat[i],sizeof(int)*max);//初始化所有数字为零 保证无脑乘除法的过程中没有乱数字的情况.
cat[0][0]=1;
cat[1][0]=1;
for(int i=2;i<max;i++)
{
memcpy(cat[i],cat[i-1],sizeof(int)*max);//按照字节复制 保证数据值不变.
big_mul(cat[i],4*i-2);
big_div(cat[i],i+1);
}
}
void output(int *a,int n)
{
int i=n-1;
while(a[i]==0)i--;
printf("%d",a[i--]);
while(i>=0)printf("%04d",&t))
{
output(cat[t],max);
}
}
版权声明：本文内容由互联网用户自发贡献，该文观点与技术仅代表作者本人。本站仅提供信息存储空间服务，不拥有所有权，不承担相关法律责任。如发现本站有涉嫌侵权/违法违规的内容， 请发送邮件至 dio@foxmail.com 举报，一经查实，本站将立刻删除。
