HDU_1028 Ignatius and the Princess III 【母函数的应用之整数拆分】

题目：

"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!" 

InputThe 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. 
OutputFor 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

题意分析：

这题是对母函数的另一个应用，整数的拆分。

我们可以把每个数的数值当作母函数经典例题中的砝码的质量。然后把需要凑的总数值当作砝码需要称的质量，这题就比较好理解了。

打表，控制指数在120以内。

AC代码：

#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;
const int MAXN = 120;
int C1[MAXN+3], C2[MAXN+3];

void solve()
{
    int i, j, k;
    for(i = 0; i <= MAXN; i++)
    {
        C1[i] = 1;
        C2[i] = 0;
    }
    for(i = 2; i <= MAXN; i++)
    {
        for(j = 0; j <= MAXN; j++)
        {
            for(k = 0; k+j <= MAXN; k+=i)
            {
                C2[k+j] += C1[j];
            }
        }
        for(j = 0; j <= MAXN; j++)
        {
            C1[j] = C2[j];
            C2[j] = 0;
        }
    }
}

int main()
{
    int N;
    solve();
    while(scanf("%d", &N)!=EOF)
    {
        printf("%d\n", C1[N]);
    }
    return 0;
}