[AHOI 2005]COMMON 约数研究

Posted 2020-10-14 NaVi_Awson

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...

 1 //It is made by Awson on 2017.11.10
 2 #include <map>
 3 #include <set>
 4 #include <cmath>
 5 #include <ctime>
 6 #include <queue>
 7 #include <stack>
 8 #include <cstdio>
 9 #include <string>
10 #include <vector>
11 #include <cstdlib>
12 #include <cstring>
13 #include <iostream>
14 #include <algorithm>
15 #define LL long long
16 #define LD long double
17 #define Max(a, b) ((a) > (b) ? (a) : (b))
18 #define Min(a, b) ((a) < (b) ? (a) : (b))
19 #define sqr(x) ((x)*(x))
20 #define y1 yy
21 #define count COUNT
22 #define Lr(o) (o<<1)
23 #define Rr(o) (o<<1|1)
24 using namespace std;
25 const int N = 1000000;
26 
27 int n;
28 LL ans;
29 
30 void work() {
31     scanf("%d", &n);
32     for (int i = 1; i <= n; i++)
33     ans += n/i;
34     printf("%lld\n", ans);
35 }
36 int main() {
37     work();
38     return 0;
39 }