hdu 1028 Ignatius and the Princess III

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：

#include<bits/stdc++.h>
using namespace std;
const int _max = 10001; 
// c1是保存各项质量砝码可以组合的数目
// c2是中间量，保存每一次的情况
int c1[_max], c2[_max];   
int main()
{   //int n,i,j,k;
    int nNum;   // 
    int i, j, k;
　　 while(cin >> nNum)
    {
        for(i=0; i<=nNum; ++i)   // ---- ①
        {
            c1[i] = 1;
            c2[i] = 0;
        }
        for(i=2; i<=nNum; ++i)   // ----- ②
        {
　　　　　　　　for(j=0; j<=nNum; ++j)   // ----- ③
                for(k=0; k+j<=nNum; k+=i)  // ---- ④
                {
                    c2[j+k] += c1[j];
                }
            for(j=0; j<=nNum; ++j)     // ---- ⑤
            {
                c1[j] = c2[j];
                c2[j] = 0;
            }
        }
        cout << c1[nNum] << endl;
    }
    return 0;
}

本题代码：

#include<bits/stdc++.h>
using namespace std;
#define ll long long

int c1[130], c2[130];
int main()
{
    int nNum;
    while(cin >> nNum)
    {
        // 初始化
        for(int i=0; i<=nNum; ++i)
        {
            c1[i] = 1;
            c2[i] = 0;
        }
        for(int i=2; i<=nNum; ++i)
        {
            for(int j=0; j<=nNum; ++j)
                for(int k=0; k+j<=nNum; k+=i)
                    c2[k+j] += c1[j];
            for(int j=0; j<=nNum; ++j)
            {
                c1[j] = c2[j];
                c2[j] = 0;
            }
        }
        printf("%d\n", c1[nNum]);
    }
    return 0;
}