hdu 1028 整数的划分问题

Posted 2021-03-02 fengzeng666

"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

 

 

思路

假设partition(n,m)：正整数n的划分中加数小于等于m的所有划分数。

一般情况下，有：

partion(n, m) = partition(n, m-1) + partition(n-m, m);

 

例如：

比如partition(7,4) = partition(7,3)+partition(3,4)，为什么会加上partition(3,4)呢？

4+3, 4+2+1，4+1+1+1这一行的总个数就是partition(3,4)，

相当于把最前面固定的4去掉，剩余项的和等于7-4=3的总个数，但同时剩余项的加数要小于4，因为这些数排在删掉的4的后面。

 

特殊情况(递归终止条件)：

（1）partition(n,m) = partition(n,n-1) + 1

         n<=m时，有 partition(n,m) = partition(n,n-1) + 1，因为n的划分不能有大于n的加数

（2）partition(1,n) = 1

         n = 1时，不管m有多大，整数1都只有1个划分

（3）partition(n,1) = 1

         对任意整数n，加数小于等于1的划分只有1个，即1+1+....+1

 

使用记忆化搜索优化递归次数，得到如下代码：

 1 #include <iostream>
 2 #include <vector>
 3 #include <stdio.h>
 4 #include <string>
 5 
 6 using namespace std;
 7 
 8 #define MAXN 150
 9 
10 vector<vector<int>> memo;
11 
12 int partition(int n, int m)
13 {
14     if(n < 1 || m < 1)
15         return 0;
16     
17     if(n == 1 || m == 1)
18         return 1;
19     
20     if(memo[n][m] != -1)
21         return memo[n][m];
22         
23     if(n <= m)
24         memo[n][m] = 1+partition(n,n-1);
25     else
26         memo[n][m] = partition(n,m-1) + partition(n-m,m);
27     
28     return memo[n][m];
29     
30 } 
31 
32 int main()
33 {
34     int n;
35     while(scanf("%d", &n) != EOF)
36     {
37         memo = vector<vector<int>>(MAXN, vector<int>(MAXN, -1));
38         printf("%d
", partition(n,n));
39     }
40     
41     return 0;
42 }