标签：dem ror src 使用 workspace int hat 删掉 ring
"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\n", partition(n,n));
39 }
40
41 return 0;
42 }
hdu 1028 整数的划分问题
标签：dem ror src 使用 workspace int hat 删掉 ring
原文地址：https://www.cnblogs.com/FengZeng666/p/12663187.html
