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#include <bits/stdc++.h>
using namespace std;
int n;
const int N = 1010;
int f[N];
int dfs(int x) {
//存在就返回
if (f[x]) return f[x];
//1就没法继续分了，同时，由于题目说：原数列不做任何修改就直接统计为一种合法数列。所以返回1
if (x == 1) return f[x] = 1;
//不是1，可以分
int ans = 1;//它自己就是一种
//在它后面不断加入[1,x/2]的数字，都可以增加方法数量
for (int i = 1; i <= x / 2; i++) ans += dfs(i);
//返回方法数量
return f[x] = ans;
}
int main() {
//输入
cin >> n;
//计算并输出
cout << dfs(n) << endl;
return 0;
