hdu 1028 整数的划分问题

"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：

#include <iostream>
#include <vector>
#include <stdio.h>
#include <string>

using namespace std;

#define MAXN 150

vector<vector<int>> memo;

int partition(int n, int m)
{
    if(n < 1 || m < 1)
        return 0;
    
    if(n == 1 || m == 1)
        return 1;
    
    if(memo[n][m] != -1)
        return memo[n][m];
        
    if(n <= m)
        memo[n][m] = 1+partition(n,n-1);
    else
        memo[n][m] = partition(n,m-1) + partition(n-m,m);
    
    return memo[n][m];
    
} 

int main()
{
    int n;
    while(scanf("%d", &n) != EOF)
    {
        memo = vector<vector<int>>(MAXN, vector<int>(MAXN, -1));
        printf("%d\n", partition(n,n));
    }
    
    return 0;
}