ural 1017. Staircases DP

最新推荐文章于 2019-07-22 13:02:26 发布

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本文介绍了一个有趣的问题：如何使用一定数量的小砖块构建不同类型的楼梯状排列，楼梯必须由大小递减的步骤组成且每一步至少包含一块砖。文章提供了一个Java程序实现，通过动态规划方法来计算可以构建的不同楼梯状排列的数量。

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1017. Staircases

Time limit: 1.0 second
Memory limit: 64 MB

One curious child has a set of N little bricks (5 ≤ N ≤ 500). From these bricks he builds different staircases. Staircase consists of steps of different sizes in a strictly descending order. It is not allowed for staircase to have steps equal sizes. Every staircase consists of at least two steps and each step contains at least one brick. Picture gives examples of staircase for N=11 andN=5:

Your task is to write a program that reads the number N and writes the only number Q — amount of different staircases that can be built from exactly N bricks.

Input

Number N

Output

Number Q

Sample

input	output
212	995645335

Problem Source: Ural State University Internal Contest '99 #2

import java.io.BufferedReader;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.math.BigInteger;
import java.util.Arrays;
import java.util.StringTokenizer;

public class Main {

	public static void main(String[] args) {
		new Task().solve();
	}
}

class Task {
	InputReader in = new InputReader(System.in) ;
	PrintWriter out = new PrintWriter(System.out);
	
	long[][] dp = new long[501][501] ;
	
	long dfs(int n , int m){
		if(n < 0 || m < 0){
			return 0 ;		
		}
		if(n == 0){
			return 1 ;
		}
		if(dp[n][m] != -1){
			return dp[n][m] ;
		}
		return dp[n][m] = dfs(n-m, m-1) + dfs(n,m-1) ;
	}

	void solve() {
		for(int i = 0 ; i <= 500 ; i++){
			Arrays.fill(dp[i] , -1) ;
		}
		int n = in.nextInt() ;
		out.println(dfs(n , n )-1) ;
		out.flush();
	}
	
}


class InputReader {  
    public BufferedReader reader;  
    public StringTokenizer tokenizer;  
  
    public InputReader(InputStream stream) {  
        reader = new BufferedReader(new InputStreamReader(stream), 32768);  
        tokenizer = new StringTokenizer("");  
    }  
  
    private void eat(String s) {  
        tokenizer = new StringTokenizer(s);  
    }  
  
    public String nextLine() {  
        try {  
            return reader.readLine();  
        } catch (Exception e) {  
            return null;  
        }  
    }  
  
    public boolean hasNext() {  
        while (!tokenizer.hasMoreTokens()) {  
            String s = nextLine();  
            if (s == null)  
                return false;  
            eat(s);  
        }  
        return true;  
    }  
  
    public String next() {  
        hasNext();  
        return tokenizer.nextToken();  
    }  
  
    public int nextInt() {  
        return Integer.parseInt(next());  
    }  
      
    public int[] nextInts(int n){  
        int[] nums = new int[n] ;  
        for(int i = 0 ; i < n ; i++){  
            nums[i] = nextInt() ;  
        }  
        return nums ;  
    }  
  
    public long nextLong() {  
        return Long.parseLong(next());  
    }  
  
    public double nextDouble() {  
        return Double.parseDouble(next());  
    }  
  
    public BigInteger nextBigInteger() {  
        return new BigInteger(next());  
    }  
  
}