HDU 1028（母函数）

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Ignatius and the Princess III

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 19181    Accepted Submission(s): 13477

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

题意：给你一个正整数n，将该正整数拆分成若干整数的和，问你共有多少不同的拆分方法？

#include<bits/stdc++.h>
using namespace std;
int c1[],c2[];//c1用于记录各项前面的系数，c2用于记录中间值
int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        //共有n个括号，先处理第一个括号，第一个括号里每一项的系数都是1；
        for(int i=;i<=n;i++)
        {
            c1[i]=;
            c2[i]=;
        }
        //因为共有n个括号，i表示正在处理第i个括号
        for(int i=;i<=n;i++)
        {
            //j表示正在处理第i个括号里的第就项，
            for(int j=;j<=(n);j++)
            {
                for(int k=;k+j<=n;k+=i)
                {
                    c2[j+k]+=c1[j];
                }
            }
            for(int j=;j<=(n);j++)
            {
                c1[j]=c2[j];
                c2[j]=;
            }
        }
        printf("%d\n",c1[n]);
    }
    return ;
}