"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

代码：

/*  gyt
       Live up to every day            */

#include<cstdio>
#include<cmath>
#include<iostream>
#include<algorithm>
#include<vector>
#include<stack>
#include<cstring>`
#include<queue>
#include<set>
#include<string>
#include<map>
#include <time.h>
#define PI acos(-1)
using namespace std;
typedef long long ll;
typedef double db;
const int maxn = 10000+5;
const ll maxm = 1e7;
const ll mod = 1e9 + 7;
const int INF = 0x3f3f3f;
const ll inf = 1e15 + 5;
const db eps = 1e-9;
int c1[maxn], c2[maxn];

void solve() {
    int n;
    while(scanf("%d", &n)!=EOF) {
        for (int i=0; i<=n; i++) {
            c1[i]=1, c2[i]=0;
        }
        for (int i=2; i<=n; i++) {
            for (int j=0; j<=n; j++) {
                for (int k=0; j+k<=n; k+=i) {
                    c2[j+k]+=c1[j];
                }
            }
            for (int j=0; j<=n; j++) {
                c1[j]=c2[j];
                c2[j]=0;
            }
        }
        printf("%d\n", c1[n]);
    }
}
int main() {
    int t = 1;
    //freopen("in.txt", "r", stdin);
    //scanf("%d", &t);
    while(t--)
        solve();
    return 0;
}