杭电1023 Train problemII(卡塔兰大数)

Train Problem II

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5923    Accepted Submission(s): 3219

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.

 

Author

Ignatius.L

/*
Time:2014-11-28 20:30 更新
*/
//卡特兰大数
//f[n]=f[n-1]*(4*n-2)/(n+1) 1,2,5,14
#include
#include
#include
using namespace std;
const int MAX=1000;
int f[105][MAX];
int digit[MAX];
void Catalan(){
	memset(f,0,sizeof(f));
	f[0][0]=1;f[1][0]=1;
	digit[0]=digit[1]=0;
	int i,j,t;
	for(i=2;i<=100;i++){
		int t=0;digit[i]=digit[i-1];
		for(j=0;j<=digit[i];j++){
			f[i][j]=f[i-1][j]*(4*i-2)+t;
			t=f[i][j]/10;
			f[i][j]%=10;
			if(t&&j+2>=digit[i]){
				digit[i]++;
			}
		}/*
		for(j=digit[i];j>=0;j--){
			printf("%d",f[i][j]);
		}puts("");*/
		t=0;
		for(j=digit[i];j>=0;j--){
			f[i][j]=f[i][j]+t*10;
			t=f[i][j]%(i+1);
			f[i][j]/=(i+1);
		}
		while(f[i][digit[i]]==0)digit[i]--;
	}
} 
int main(){
	int i,j,n;
	Catalan();
	while(scanf("%d",&n)!=EOF){
		for(i=digit[n];i>=0;i--){
			printf("%d",f[n][i]);
		}puts("");
		//printf("%d\n",digit[n]);
		
	}
return 0;
}

/*
被大数吓怕了，卡特兰大数，因为是大数与小数的运算，所以也不是很难，看见包子早就做了，我也做过，不过64位也wa知道是大数就不敢碰了  加油！！！
Time:2014-10-3 17:54
*/
#include//卡特兰数公式f[n]=f[n-1]*(4n-2)/(n+1) 
#include
#define MAX 220
int a[MAX][MAX],b[MAX];//b数组来记录每个数的位数 
void Init(){
	a[1][0]=1;b[1]=0;
	int i,j,k=1;//k表示位数
	int z; 
	for(i=2;i<102;i++){
		for(j=0;j=0;j--){
		 	a[i][j]=a[i][j]+z*10;
		 	z=a[i][j]%(i+1);
		 	a[i][j]/=(i+1);
		 }
		 while(a[i][k-1]==0)k--;//注意：是a[][k-1] 不是k
		 b[i]=k-1; 
	} 
}
int main(){
	int N;
	Init();
	while(scanf("%d",&N)!=EOF){
		for(int i=b[N];i>=0;i--)
		printf("%d",a[N][i]);
		puts("");
	}
return 0;
}