约数和

简单来讲，就是要求$\sum_{i=1}^n(\sum_{d|i}d)$

$$\sum_{i=1}^n(\sum_{d|i}d)=\sum_{i=1}^n(\sum_{j=1}^n{[j|i]*j})=\sum_{j=1}^n(\sum_{i=1}^n{[j|i]*j})=\sum_{j=1}^n(j*{\lfloor}{\frac{n}{j}}{\rfloor})$$

$[j|i]$表示如果j是i的因子则为1，否则为0

（sigma性质？https://wenku.baidu.com/view/1006651014791711cc79171f.html）

所以就跟https://www.cnblogs.com/hehe54321/p/7307069.html一样了。。。

题面就有很多了。。。

---

https://www.luogu.org/problemnew/show/P2424

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<vector>
using namespace std;
#define fi first
#define se second
#define mp make_pair
#define pb push_back
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> pii;
ll x,y,ans;
int main()
{
    ll i,j;
    scanf("%lld%lld",&x,&y);
    for(i=1;i<=y;i=j+1)
    {
        j=min(y,y/(y/i));
        ans+=(y/i)*(i+j)*(j-i+1)/2;
    }
    x--;
    for(i=1;i<=x;i=j+1)
    {
        j=min(x,x/(x/i));
        ans-=(x/i)*(i+j)*(j-i+1)/2;
    }
    printf("%lld",ans);
    return 0;
}

---

http://210.33.19.103/problem/3415

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<vector>
using namespace std;
#define fi first
#define se second
#define mp make_pair
#define pb push_back
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> pii;
ll y,ans;
int main()
{
    ll i,j;
    scanf("%lld",&y);
    if(y==0)
    {
        puts("0");
        return 0;
    }
    for(i=1;i<=y;i=j+1)
    {
        j=min(y,y/(y/i));
        ans+=(y/i)*(i+j)*(j-i+1)/2;
    }
    ans-=y;ans-=(y+1)*y/2;ans++;
    printf("%lld",ans);
    return 0;
}