hdu 1028 Ignatius and the Princess III(母函数)-优快云博客

本文探讨了一个经典的数学问题——整数划分。对于给定的正整数N，我们需要找出所有可能的不同整数组合方式，使得这些整数之和等于N。文章通过一个具体的例子进行了说明，并提供了一段C++代码实现，用于解决该问题。

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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

41020

Sample Output

542627

Author

Ignatius.L

#include<stdio.h>
#include<algorithm>
#include<string.h>
#include<iostream>
#include<math.h>
#include<queue>
#include<map>
#include<vector>
#include<set>
using namespace std;
#define inf 0x3f3f3f3f
#define ll long long
int c1[130], c2[130];
int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        for(int i=0; i<=n; ++i)
        {
            c1[i] = 1;
            c2[i] = 0;
        }
        for(int i=2; i<=n; ++i)
        {
            for(int j=0; j<=n; ++j)
                for(int k=0; k+j<=n; k+=i)
                {
                    c2[k+j] += c1[j];
                }
            for(int j=0; j<=n; ++j)
            {
                c1[j] = c2[j];
                c2[j] = 0;
            }
        }
        printf("%d\n", c1[n]);
    }
    return 0;
}