對於數字i,在1到n中,一共有n / i個數是i的倍數。
/* Think Thank Thunk */
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
inline int iread() {
int f = 1, x = 0; char ch = getchar();
for(; ch < '0' || ch > '9'; ch = getchar()) f = ch == '-' ? -1 : 1;
for(; ch >= '0' && ch <= '9'; ch = getchar()) x = x * 10 + ch - '0';
return f * x;
}
int main() {
int n = iread(), ans = 0;
for(int i = 1; i <= n; i++) ans += n / i;
printf("%d\n", ans);
return 0;
}