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#include <bits/stdc++.h>
using namespace std;
int n;
//毫不意外只通过了5个测试点TLE了15个点~
int dfs(int 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 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 ans;
}
int main() {
//输入
cin >> n;
//计算并输出
cout << dfs(n) << endl;
return 0;
}