版权声明:本文为博主瞎写的,请随便转载 https://blog.csdn.net/sdut_jk17_zhangming/article/details/82792755
LL get_sum(LL n){
LL ans = 0;
for(LL l = 1,r;l <= n;l = r + 1){
r = n / (n / l);
ans += (n / l) * (l + r) * (r - l + 1) / 2;
}
return ans;
}