hoj1294

用dfs进行整数拆分，对于每种情况计算组合数。为了避免重复计算，必须使用有重集的组合，com(f[i] - 1 + num[i], num[i])，这是num[i]个有i个节点的子树的情况总数。这个公式是用于求所选物品可重复选择，且重复选出的同一物品完全相同的情况。

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
using namespace std;

#define maxn 41

long long f[maxn], num[maxn], ans;
int n;

long long com(long long n, long long r)
{// return C(n, r)
    if (n - r < r)
        r = n - r; // C(n, r) = C(n, n-r)
    long long i, j, s = 1;
    for (i = 0, j = 1; i < r; ++i)
    {
        s *= (n - i);
        for (; j <= r && s % j == 0; ++j)
            s /= j;
    }
    return s;
}
long long gcd(long long a, long long b)
{
    return b?gcd(b,a%b):a;
}

long long C(long long n, long long m)
{
    if(m > n - m ) m = n - m;
    long long mul = 1 , div = 1;
    for(int i = 0; i < m;i++)
    {
        mul *= (n-i);
        div *= (i+1);
        long long g = gcd(mul,div);
        mul/=g , div /= g;
    }
    return mul / div;
}


void dfs(int s, int left)
{
    if (left == 0)
    {
        ans = 1;
        for (int i = 0; i <= n; i++)
            if (num[i] != 0)
                ans *= com(f[i] - 1 + num[i], num[i]);
        f[n] += ans;
    }
    if (s > left)
        return;
    num[s]++;
    dfs(s, left - s);
    num[s]--;
    dfs(s + 1, left);
}

int main()
{
    //freopen("D:\\t.txt", "r", stdin);
    //freopen("D:\\y.txt", "w", stdout);
    memset(num, 0, sizeof(num));
    f[1] = 1;
    f[2] = 1;
    for (n = 3; n <= 40; n++)
    {
        dfs(1, n - 1);
        //printf("f[%d]=%I64d;", n, f[n]);
    }
    while (scanf("%d", &n) != EOF)
        printf("%I64d\n", f[n]);
    return 0;
}