HDU1028 Ignatius and the Princess III（java）

Problem Description
“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.

Sample Input
4
10
20

Sample Output
5
42
627

題意是將一個整數N分解爲不大於N的各個整數和的形式的個數。
這裏使用動態規劃來完成。dp[i][j]爲將i分解爲不大於j的個數。

import java.util.Arrays;
import java.util.Scanner;


public class P1027 {
    int[][] dp = new int[130][130];

    public static void main(String[] args) {
        new P1027().run();
    }

    public void run(){
        Scanner scanner = new Scanner(System.in);
        int n;
        while (scanner.hasNextInt()) {
            n = scanner.nextInt();
            for (int i = 0; i <= n; i++) {
                Arrays.fill(dp[i], -1);
            }
            System.out.println(calc(n, n));
        }
        scanner.close();
    }

    //dp[i][j]表示 i 表示成最大的數不超過 j 的方法數
    int calc(int n, int m){
        if(dp[n][m] != -1)
            return dp[n][m];
        if(n < 1 || m < 1)
            return dp[n][m] = 0;
        if(n == 1 || m ==1)
            return dp[n][m] = 1;
        if(n < m)
            return dp[n][m] = calc(n, n);
        if(n == m)
            return dp[n][m] = calc(n, m - 1) + 1;
        //dp[n][m]由不包括m的數字組成的n和包括m組成n的個數
        return dp[n][m] = calc(n, m - 1) + calc(n - m, m);
    }
}

注：提交到OJ上，請將主類名改爲Main。