Ignatius and the Princess III

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 13129    Accepted Submission(s): 9285

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

Author

Ignatius.L

Mean:

给你一个n，问你n的拆分有多少种。

analyse:

经典的母函数运用题，不解释。

Time complexity：O(n^3)

Source code：

// Memory   Time
// 1347K     0MS
// by : Snarl_jsb
// 2014-09-18-15.21
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<vector>
#include<queue>
#include<stack>
#include<map>
#include<string>
#include<climits>
#include<cmath>
#define N 1000010
#define LL long long
using namespace std;
int c1[300],c2[300];
int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(0);
//    freopen("C:\\Users\\ASUS\\Desktop\\cin.cpp","r",stdin);
//    freopen("C:\\Users\\ASUS\\Desktop\\cout.cpp","w",stdout);
    int n;
    while(cin>>n)
    {
        memset(c1,0,sizeof(c1));
        memset(c2,0,sizeof(c2));
        for(int i=0;i<=n;++i)
        {
            c1[i]=1;
        }
        for(int i=2;i<=n;i++)
        {
            for(int j=0;j<=n;++j)
            {
                for(int k=0;k<=n;k+=i)
                {
                    c2[k+j]+=c1[j];
                }
            }
            for(int j=0;j<=n;++j)
            {
                c1[j]=c2[j];
                c2[j]=0;
            }
        }
        cout<<c1[n]<<endl;
    }
    return 0;
}