#include <cstdio>#include <cstring>#include <cmath>#include <iostream>#include <algorithm>using namespace std;#define LL __int64const LL maxn=44;LL dp[maxn]={0,1,1,2,4,9,20,48,115,286,719,1842,4766,12486,32973,87811,235381,634847,1721159,4688676, 12826228,35221832,97055181,268282855,743724984,2067174645,5759636510,16083734329, 45007066269,126186554308,354426847597,997171512998,2809934352700,7929819784355, 22409533673568,63411730258053,179655930440464,509588049810620,1447023384581029, 4113254119923150,11703780079612453};//可以直接打表了。。LL e[maxn],t,s;LL C(LL n,LL m)//求组合数{ m=min(m,n-m); LL i,s=1; for(i=1;i<=m;i++) s=s*(n-i+1)/i; return s;}void dfs(LL a,LL step,LL num,LL sum){ LL i,j,k; if(sum>num)return; if(sum==num) { LL s=1; k=1; for(i=1;i<step;i++) { if(e[i]!=e[i-1]) { s*=C(dp[e[i-1]]+k-1,k); k=0; } k++; } s*=C(dp[e[step-1]]+k-1,k); dp[num+1]+=s; return; } for(i=a;i<=num;i++) { e[step]=i; dfs(i,step+1,num,sum+i); }}void init(){ dp[1]=dp[2]=1; LL i,j,k; for(i=3;i<=40;i++) { dp[i]=0; dfs(1,0,i-1,0); }}int main(){ init(); LL n; while(cin>>n) { cout<<dp[n]<<endl; } return 0;}/* 可重复选择的组合。有n个元素，每个元素可以选多次，一共选k个袁术，有多少种选择？ n=3,k=2,有6种:(1,1),(1,2),(1,3),(2,2),(2,3),(3,3) 共 C(k+n-1,n-1)=C(n+k-1,k)种 本题，求n个结点数的树有多少种，可采用整数划分n-1的方法 n=5时，划分4=1+1+1+1=1+1+2=1+3=2+2=4。 则dp[5]=C(dp[1]+4-1,4)+C(dp[1]+2-1,2)*dp[2]+dp[1]*dp[3]+dp[2]*dp[2]+dp[4]=9;*/
