杭电ACM OJ 1028 Ignatius and the Princess III 母函数+优化

Ignatius and the Princess III

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 23133    Accepted Submission(s): 16129

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

41020

 

Sample Output

542
627
翻译：
  4 = 4;
  4 = 3 + 1;
  4 = 2 + 2;
  4 = 2 + 1 + 1;
  4 = 1 + 1 + 1 + 1;
例如就是把4的所有加出来的情况都列出来，看看一共有几种。
思路：
粗粗看了一眼，动态规划可以做。不过显然，用母函数是更为优雅，标准的做法。
想深刻学习下母函数的，可以看
http://blog.csdn.net/qq_36523667/article/details/78682444
用两个经典例子+一个附加例子把母函数讲的十分透彻，可以说是目前最易于理解的母函数的文章了。
本题母函数思路：
就是上面文章里的最后一题。我直接拷一下思路。
[代码思路：
传进来的是n，开n个数组即可。就好比这个和为5的这个例子
数组1：0 1 2 3 4 5
数组2：0 2 4
数组3：0 3
数组4：0 4
数组5：0 5

就解决了。题外话，难道n个数组就傻傻地开n个for循环吗，最后我看了一下。开3重循环，从最后一层开始找就很快了。这样一来时间复杂度远没有O(n^3)。]
代码：(就是弄一个栈不断地去加每一行的值，并保存下来,从最后一行开始是效率最高的)
package ACM1000_1099;// TODO: 2017/12/1 母函数+优化import java.util.ArrayList;import java.util.List;import java.util.Map;public class IgnatiusAndThePrincessThree1028 {    IgnatiusAndThePrincessThree1028(int n) {        calculate(n);    }    int calculate(int n) {        int[][] a = new int[n][n + 1];        //给n个数组赋值        for (int row = 0; row < n; row++) {            int index = row + 1;//倍数            for (int col = 0; col < n + 1; col++) {                int num = col * index;                if (num <= n) {                    a[row][col] = num;                }            }        }//        for (int row = 0; row < n; row++) {//            for (int col = 0; col < n + 1; col++) {//                System.out.print(a[row][col] + " ");//            }//            System.out.println();//        }        List<Integer> list = new ArrayList<>();        list.add(0);        List<Integer> tempList = new ArrayList<>();//        for (int col1 = 0; col1 < n + 1; col1 ++) {//            if (col1 != 0 && a[4][col1] == 0) {//                break;//            } else {//                for (int col2 = 0; col2 < n + 1; col2 ++) {//                    if (col2 != 0 && a[3][col2] == 0) {//                        break;//                    } else {//                        int num = a[4][col1] + a[3][col2];//                        if (num <= n) {//                            list.add(num);//                        }//                    }//                }//            }//        }        for (int row = 0; row < n; row ++) {            for (int col = 0; col < n + 1; col ++) {                if (col != 0 && a[row][col] == 0) {                    break;                } else {                    for (int i = 0; i < list.size(); i ++) {                        int num = list.get(i) + a[row][col];                        if (num <= n) {                            tempList.add(num);                        }                    }                }            }            list.clear();            list.addAll(tempList);            tempList.clear();        }        int result = 0;        for (int i = 0; i < list.size(); i ++) {            if (list.get(i) == n) {                result ++;            }        }        System.out.println(result + "");        return 0;    }    public static void main(String[] args) throws Exception {        IgnatiusAndThePrincessThree1028 i = new IgnatiusAndThePrincessThree1028(20);    }}