整数划分问题

/**
*@ author StormMaybin
*@ date 2016-09-19
*/

生命不息，奋斗不止！

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HDOJ 1028

题目描述
“Well, it seems the first problem is too easy. I will let you know how foolish you are later.” feng5166 says.
“The second problem is, given an positive integer N, we define an equation like this:
N=a[1]+a[2]+a[3]+…+a[m];
a[i]>0,1<=m<=N;
My question is how many different equations you can find for a given N.
For example, assume N is 4, we can find:
4 = 4;
4 = 3 + 1;
4 = 2 + 2;
4 = 2 + 1 + 1;
4 = 1 + 1 + 1 + 1;
so the result is 5 when N is 4. Note that “4 = 3 + 1” and “4 = 1 + 3” is the same in this problem. Now, you do it!”

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Input
The input contains several test cases. Each test case contains a positive integer N(1<=N<=120) which is mentioned above. The input is terminated by the end of file.

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Output
For each test case, you have to output a line contains an integer P which indicate the different equations you have found.

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Sample Input
4
10
20

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Simple Output
5
42
627

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题目分析

假设dp[i][j] 代表把i划分成不大于j的一组数的和的种数，可以得知dp[i][1] = dp[1][i] = 1。如果i < j那么dp[i][j] = dp[i][i]，如果i > j，dp[i][j] = dp[i-j][j] + dp[i][j-1]，如果i == j，那么dp[i][j] = 1 + dp[i][j-1]。至于为什么这样，稍微动脑想一下应该很容易能懂，沃野不动手写过程了，心累，更完睡觉…….

演示代码

package com.stormma.dp;

import java.io.BufferedInputStream;
import java.util.Scanner;

public class Main1028
{

    /**
     * @param args
     */
    private Scanner scan;
    public Main1028()
    {
        scan = new Scanner(new BufferedInputStream(System.in));
        while (scan.hasNext())
        {
            int n = scan.nextInt();
            int [][] dp = new int [n+1][n+1];
            //dp[i][j]表示把i分成不大于j的数的种数。
            dp[n][1] = 1;
            init(dp, n);
            for(int i=2; i <= n; i++)
            {
                for(int j=2; j <= n; j++)
                {
                    if(i < j) 
                        dp[i][j] = dp[i][i];
                    else if(i == j)
                        dp[i][j]= 1+dp[i][j-1];
                    else if(i > j) 
                        dp[i][j]=dp[i-j][j]+dp[i][j-1];
                }
            }
            System.out.println(dp[n][n]);
        }
    }
    private void init(int [][] dp, int n)
    {
        for (int i = 1; i <= n; i++)
            dp[1][i] = dp[i][1] = 1;
    }
    public static void main(String[] args)
    {
        // TODO Auto-generated method stub
        new Main1028();
    }

}