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#include <bits/stdc++.h>
using namespace std;
const int N = 1010;
int v[5] = {0, 10, 20, 50, 100};
int f[N];
// 体积限制是恰好是，因此需要初始化f[0][0]为合法解1，其他位置为非法解0。
int main() {
int m;
cin >> m;
// 前0种物品，体积是0的情况下只有一种方案
f[0] = 1;
for (int i = 1; i <= 4; i++)
for (int j = v[i]; j <= m; j++)
f[j] += f[j - v[i]];
// 输出
printf("%d\n", f[m]);
return 0;
}
