求前n个正整数的约数之和

求\(\sum_{i=1}^{n}\sigma \left ( i\right ),i\leq 10^{12},\sigma \left ( n\right )=\sum_{d|n}d\)

解：\(\sum_{i=1}^{n}\sigma \left ( i\right )=\sum_{i=1}^{n}\sum_{d|i}d=\sum_{i=1}^{n}\sum_{d=1}^{n}\left [ d|i\right ]*d=\sum_{i=1}^{n}i*\sum_{i=1}^{n}\left [ i|d\right ]=\sum_{i=1}^{n}i*\left \lfloor \frac{n}{i}\right \rfloor\)

代码：

ll n,ans;
int main()
{
    scanf("%lld",&n);
    for(ll l=1ll,r;l<=n;l=r+1ll){
        r=min(n/(n/l),n);
        ans = ans + (n/l)*((l+r)*(r-l+1)/2);
    }
    printf("%lld\n",ans);
    return 0;
}

也可以这样化简：

\(\sum_{i=1}^{n}\sum_{d|n}d=\sum_{i=1}^{n}\sum_{d=1}^{\left \lfloor \frac{n}{i}\right \rfloor}d=\sum_{i=1}^{n}\frac{\left ( 1+\left \lfloor \frac{n}{i}\right \rfloor\right )*\left \lfloor \frac{n}{i}\right \rfloor}{2}\)

\(\sum_{i=1}^{n}\frac{\left ( 1+\left \lfloor \frac{n}{i}\right \rfloor\right )*\left \lfloor \frac{n}{i}\right \rfloor}{2}=\sum_{i=1}^{n}i*\left \lfloor \frac{n}{i}\right \rfloor\)