hdu 1028 Ignatius and the Princess III

Posted 2020-06-18 lucky_少哖

Ignatius and the Princess III

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 16918    Accepted Submission(s): 11907

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

 

 

Author

Ignatius.L

 

 

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整数划分问题，和我前面写的一篇博客是一样的，参考：http://www.cnblogs.com/pshw/p/4838898.html

不过此题数据比较大，直接递归计算会导致超时，因此需要使用母函数的思想。其实我也不知道啥叫母函数。。。只是用dp将数据保存起来。

 

题意：整数划分问题是将一个正整数n拆成一组数连加并等于n的形式，且这组数中的最大加数不大于n。

 

附上代码：

 

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 using namespace std;
 5 int main()
 6 {
 7     int n,i,j,m;
 8     int dp[125][125];
 9     memset(dp,0,sizeof(dp));
10     dp[1][1]=1;
11     for(i=1; i<=120; i++)
12     {
13         dp[i][1]=1;
14         dp[1][i]=1;
15     }
16     for(i=2; i<=120; i++)  //规律可见推荐的博客~
17     {
18         for(j=2; j<=120; j++)
19         {
20             if(j>i)
21                 dp[i][j]=dp[i][i];
22             else if(i==j)
23                 dp[i][j]=dp[i][j-1]+1;
24             else
25                 dp[i][j]=dp[i][j-1]+dp[i-j][j];
26         }
27     }
28     while(~scanf("%d",&n))
29     {
30         printf("%d\\n",dp[n][n]);
31     }
32     return 0;
33 }