hdu 1028 Ignatius and the Princess III

Posted 2021-02-08 lu1nacy

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

 

 

这个题可以用整数划分，也可以用母函数的模板

母函数理解参考http://www.wutianqi.com/blog/596.html

母函数模板：

 1 #include<bits/stdc++.h>
 2 using namespace std;
 5 const int _max = 10001; 
 6 // c1是保存各项质量砝码可以组合的数目
 7 // c2是中间量，保存每一次的情况
 8 int c1[_max], c2[_max];   
 9 int main()
10 {   //int n,i,j,k;
11     int nNum;   // 
12     int i, j, k;
13 　　 while(cin >> nNum)
14     {
15         for(i=0; i<=nNum; ++i)   // ---- ①
16         {
17             c1[i] = 1;
18             c2[i] = 0;
19         }
20         for(i=2; i<=nNum; ++i)   // ----- ②
21         {
22 　　　　　　　　for(j=0; j<=nNum; ++j)   // ----- ③
23                 for(k=0; k+j<=nNum; k+=i)  // ---- ④
24                 {
25                     c2[j+k] += c1[j];
26                 }
27             for(j=0; j<=nNum; ++j)     // ---- ⑤
28             {
29                 c1[j] = c2[j];
30                 c2[j] = 0;
31             }
32         }
33         cout << c1[nNum] << endl;
34     }
35     return 0;
36 }

本题代码：

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 #define ll long long
 4 
 5 int c1[130], c2[130];
 6 int main()
 7 {
 8     int nNum;
 9     while(cin >> nNum)
10     {
11         // 初始化
12         for(int i=0; i<=nNum; ++i)
13         {
14             c1[i] = 1;
15             c2[i] = 0;
16         }
17         for(int i=2; i<=nNum; ++i)
18         {
19             for(int j=0; j<=nNum; ++j)
20                 for(int k=0; k+j<=nNum; k+=i)
21                     c2[k+j] += c1[j];
22             for(int j=0; j<=nNum; ++j)
23             {
24                 c1[j] = c2[j];
25                 c2[j] = 0;
26             }
27         }
28         printf("%d
", c1[nNum]);
29     }
30     return 0;
31 }