How Many Trees?

2548 ワード
数論 さしあげる 大きい整数
How Many Trees?
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 4555 Accepted Submission(s): 2534
Problem 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). 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
Source
UVA
 
ACコード:
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;
#define rep(i,a,n) for (int i=a;i=a;i--)
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
typedef vector VI;
typedef long long ll;
typedef pair PII;
typedef pair PLL;
const ll mod = 1000000007;
const int N = 5e5 + 5; //       
int n;
int a,b;
ll dp[105][444];
int main() {
	//dp1[0]=0,dp2[0]=0;
	memset(dp,0,sizeof(dp));
	dp[0][0]=1,dp[1][0]=1,dp[2][0]=2,dp[3][0]=5; 
	rep(i,4,101){
		int c=0;int j;
		for( j=0; j<444; j++){
			dp[i][j] = dp[i-1][j]*(4*i-2) + c;
			c= dp[i][j]/10; dp[i][j]%=10;
		}
		
		c=0;j=443;
		while(dp[i][j]==0) j--;
		for( ;j>=0;j--){
			if(dp[i][j]+c < (i+1)){
				c = (c+dp[i][j])*10;
				dp[i][j]=0;
			}
			else{
				int cc=dp[i][j]+c;
				dp[i][j]=cc/(i+1);
				c = cc%(i+1)*10;
			}
		}
	}
	
//	rep(i,1,101){
//		int j;
//		for(j=443;j>0;j--){
//			if(dp[i][j]) break;
//		}
//		for(;j>=0;j--){
//			cout<>n){
		int j;
		for(j=443;j>0;j--){
			if(dp[n][j]) break;
		}
		for(;j>=0;j--){
			cout<

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