hdu 1028 Ignatius and the Princess III 【整数划分】

Ignatius and the Princess III

                                                                                      Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
                                                                                                          Total Submission(s): 15730    Accepted Submission(s): 11092
    链接：  hdu 1028

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
   
   
   
   
   
   
   
   


   
   
   
   
   
   
   
   

    
    
    
    
题意
    
    
    
    
题意：赤裸裸的整数划分，求N的划分成1~N中的数的和的形式的个数。
    
    
    
    
分析
    
    
    
    
整数划分的题目有很多种解决的方法，有DP的解法，有也完全背包的解法。这里仅给出一个递归的解法，用dp[n][m]表示整数N的划分中最大的为m的方法数，那么有dp[n][m] = 
    
    
    
    
Dp[n, m]= 1;                                 (n=1 or m=1)
    
    
    
    
Dp[n, n];                                  (n<m)
    
    
    
    
1+ dp[n, m-1];                              (n=m)
    
    
    
    
Dp[n-m,m]+dp[n,m-1];                              (n>m)
    
    
    
    
 
    
    
    
    
这里链接大牛们的其他做法： 
    
    
    
    
母函数 、 递归 完全 背包 
    
    
    
    
参考代码
    
    
    
    

    
    
    
    
<pre name="code" class="cpp">#include <cmath>
#include <queue>
#include <vector>
#include <cstdio>
#include <string>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <algorithm>
using namespace std;
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#define FIN             freopen("input.txt","r",stdin)
#define FOUT            freopen("output.txt","w",stdout)
#define CASE(T)         for(scanf("%d",&T);T--;)
const int maxn = 120 + 5;
int dp[maxn][maxn];
int dfs(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] = dfs(n, n);
    if(n == m)              return dp[n][m] = dfs(n, m - 1) + 1;
    return dp[n][m] = dfs(n, m - 1) + dfs(n - m, m);
}
int main()
{
//    FIN;
    int N;
    memset(dp,-1,sizeof(dp));
    while(~scanf("%d", &N))
    {
        printf("%d\n", dfs(N, N));
    }
    return 0;
}