Train Problem II（卡特兰数+大数乘除）

Train Problem II

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 7539 Accepted Submission(s): 4062

[align=left]Problem Description[/align]
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.

[align=left]Input[/align]
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.

[align=left]Output[/align]
For each test case, you should output how many ways that all the trains can get out of the railway.

[align=left]Sample Input[/align]

1
2
3
10

[align=left]Sample Output[/align]

1
2
5
16796

Hint

The result will be very large, so you may not process it by 32-bit integers.

题解：卡特兰数：ｈ( n ) = ( ( 4*n-2 )/( n+1 )*h( n-1 ) );也可以用java，h(n)= h(0)*h(n-1) + h(1)*h(n-2) +

+ h(n-1)h(0) (其中n>=2);
这里用的大数；
代码：

#include<cstdio>
#include<iostream>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<stack>
using namespace std;
const int INF=0x3f3f3f3f;
#define mem(x,y) memset(x,y,sizeof(x))
#define SI(x) scanf("%d",&x)
#define PI(x) printf("%d",x)
int N;
//ｈ( n ) = ( ( 4*n-2 )/( n+1 )*h( n-1 ) );
int ans[110][110];
void db(){
ans[0][0]=1;
ans[0][1]=1;
ans[1][0]=1;
ans[1][1]=1;
int len=1,yu=0;
for(int i=2;i<=100;i++){
for(int j=1;j<=len;j++){
int t=ans[i-1][j]*(4*i-2)+yu;
yu=t/10;
ans[i][j]=t%10;
}
while(yu){
ans[i][++len]=yu%10;
yu/=10;
}
for(int j=len;j>=1;j--){
int t=ans[i][j]+yu*10;
ans[i][j]=t/(i+1);
yu=t%(i+1);
}
while(!ans[i][len])len--;
ans[i][0]=len;
}
}
int main(){
mem(ans,0);
db();
while(~SI(N)){
for(int i=ans
[0];i>=1;i--)printf("%d",ans
[i]);
puts("");
}
return 0;
}