HDU 1028 動態規劃

http://acm.hdu.edu.cn/showproblem.php?pid=1028

"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!"

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.

Output:

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

基本思路：

例如 對於   5  分解 爲  1 2 3 4 5 的和  可以看出是一個 容量爲 5  的揹包 對於 5 種物品  放入 剛好把揹包放滿 共有 多少中方法：（完全揹包問題）

我們 設 f[ i] [v] 前 i 個數 組合成 v 的個數  那麼 對於  f[i][ v]  = sum { f[i-1][ v- k*i]}  : 代碼如下

import java.util.Scanner;
public class HDU1028 {
	/*
	 * f[i][v] 是前 i 個數能組成的和爲 v 的個數
	 * f[i][v] = sum{ f[i-1][v-k*i] }
	 */
	private int n;
	private int f[] ;
	public void  solve(){
		Scanner sc = new Scanner(System.in);
		while(sc.hasNext()){
			n = sc.nextInt();
			f = new int[n+1];
			f[0] = 1;
			int sum =0;
			for(int i=1;i<=n; i++){
				for(int j=i;j<=n; j++){
					f[j] += f[j-i];
				}
			}
			System.out.println(f[n]);
		}
	}
	public static void main(String[] args) {
		new HDU1028().solve();
	}
}