\(\sum_{i=1}^n\;k\;mod\;i\)
Solution
\(\sum_{i=1}^n\;k\;mod\;i\\=\sum_{i=1}^n(k-i\lfloor{\frac{k}{i}}\rfloor)\\=k\times n-\sum_{i=1}^ni\lfloor{\frac{k}{i}}\rfloor\)
至于后面那项，整除分块即可
#include <bits/stdc++.h>
using namespace std;
#define int long long
signed main() {
int n,k,l=1,r,ans=0;
cin>>n>>k;
while(l<=min(n,k)) {
r=min(n,k/(k/l));
ans+=(l+r)*(r-l+1)*(k/l)/2;
l=r+1;
}
cout<<k*n-ans;
}