HDOJ 1130 How Many Trees?(卡特兰数+大数乘除法)
发布时间：2021-05-27 13:54:49 所属栏目：大数据 来源：网络整理
导读：How Many Trees? Time Limit: 2000/1000 MS (Java/Others)????Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 3380????Accepted Submission(s): 1958 Problem Description A binary search tree is a binary tree with root k such that
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How Many Trees?Time Limit: 2000/1000 MS (Java/Others)????Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 3380????Accepted Submission(s): 1958Problem Description A binary search tree is a binary tree with root k such that any node v reachable from its left has label (v) <label (k) and any node w reachable from its right has label (w) > label (k). It is a search structure which can find a node with label x in O(n log n) average time,where n is the size of the tree (number of vertices). Given a number n,can you tell how many different binary search trees may be constructed with a set of numbers of size n such that each element of the set will be associated to the label of exactly one node in a binary search tree? ? Input The input will contain a number 1 <= i <= 100 per line representing the number of elements of the set. ? Output You have to print a line in the output for each entry with the answer to the previous question. ? Sample Input 1 2 3? Sample Output 1 2 5? 题意：给出一个n，每个节点的编号为1~n，问n个节点的二叉搜索树有多少种？ 题解：和hdoj 1023一模一样啊，卡特兰数+大数。 不了解的点击：?点击打开链接? 代码如下： //h(n)=h(n-1)*(4*n-2)/(n+1);
#include<cstdio>
#include<cstring>
using namespace std;
int a[110][110];
//a[n][0]表示n的卡特兰数的长度，存储是反向的，a[n][1]表示个位数
void ktl()//打表
{
int i,j,len;
int c,s;
a[1][0]=1;
a[1][1]=1;
a[2][0]=1;
a[2][1]=2;
len=1;
for(i=3;i<101;++i)
{
c=0;
for(j=1;j<=len;++j)
{
s=a[i-1][j]*(4*i-2)+c;
c=s/10;
a[i][j]=s%10;
}
while(c)
{
a[i][++len]=c%10;
c/=10;
}
for(j=len;j>0;--j)
{
s=a[i][j]+c*10;
a[i][j]=s/(i+1);
c=s%(i+1);
}
while(!a[i][len])
len--;
a[i][0]=len;
}
}
int main()
{
ktl();
int n;
while(scanf("%d",&n)!=EOF)
{
for(int i=a[n][0];i>0;--i)
printf("%d",a[n][i]);
printf("n");
}
return 0;
}
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