//完全背包 求方案数目
//f[i][j] 只从前i个物品中选，且总体积恰好为j的方案的集合
//f[i][j]=f[i-1][j]+f[i-1][j-v*1]+f[i-1][j-v*2]+...f[i-1][j-v*s]
//f[i][j-v]= f[i-1][j-v*1]+f[i-1][j-v*2]+...f[i-1][j-v*s]
//所以 f[i][j]=f[i-1][j]+f[i][j-v]
#include <iostream>
using namespace std;
const int N = ;
int n;
int v[] = {, , , };
int f[N];
int main() {
cin >> n;
f[] = ;
for (int i = ; i < ; i ++ )
for (int j = v[i]; j <= n; j ++ )
f[j] =f[j] + f[j - v[i]];
cout << f[n] << endl;
return ;
}
