連結：https://www.luogu.org/problemnew/show/P1403
思路：1~n所有數的約數個數的和等于1的倍數個數和+2的倍數個數和+......是以f（n）=Σ(n/i)
代碼：
1 #include<bits/stdc++.h>
2 #define inf 0x3f3f3f3f
3 typedef long long ll;
4 const int M = int(1e5)*2 + 5;
5 using namespace std;
6
7 int a[M];
8
9
10 int main()
11 {
12 int n; cin >> n;
13
14 /*for (int i = 1; i <= n; i++) //這樣寫會re
15 a[i] = n / i + a[i - 1];
16 cout << a[n] << endl;*/
17 int ans = 0;
18 for (int i = 1; i <= n; i++)
19 ans += n / i;
20 cout << ans << endl;
21
22 return 0;
23 } 轉載于:https://www.cnblogs.com/harutomimori/p/10485546.html