[AHOI 2005]COMMON 约数研究

Description

Input

只有一行一个整数 N（0 < N < 1000000）。

Output

只有一行输出，为整数M，即f(1)到f(N)的累加和。

Sample Input

3

Sample Output

5

题解

水题一道...做不出的要$AFO$了...

//It is made by Awson on 2017.11.10
#include <map>
#include <set>
#include <cmath>
#include <ctime>
#include <queue>
#include <stack>
#include <cstdio>
#include <string>
#include <vector>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
#define LL long long
#define LD long double
#define Max(a, b) ((a) > (b) ? (a) : (b))
#define Min(a, b) ((a) < (b) ? (a) : (b))
#define sqr(x) ((x)*(x))
#define y1 yy
#define count COUNT
#define Lr(o) (o<<1)
#define Rr(o) (o<<1|1)
using namespace std;
const int N = 1000000;

int n;
LL ans;

void work() {
    scanf("%d", &n);
    for (int i = 1; i <= n; i++)
    ans += n/i;
    printf("%lld
", ans);
}
int main() {
    work();
    return 0;
}