"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!" 

4

10

20

5

42

627

#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;

}