Jisoo

众所周知，这个式子就是$sum_{i=1}^n(k-i*lfloorfrac{k}{i}rfloor)$

也就是$nk-sum_{i-1}^n(ilfloorfrac{k}{i}rfloor)$

右边的东西用数论分块+等差数列搞一下就可以了

数论分块的每一块的右边界是$lfloorfrac{k}{lfloorfrac{k}{i}rfloor}rfloor$

#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
long long n,k;
long long ans;
int main(){
    scanf("%d%d",&n,&k);
    ans=n*k;
    long long r;
    for(long long l=1;l<=n;l=r+1){
        if(k/l!=0) r=min(k/(k/l),n);
        else r=n;
        ans-=(k/l)*(r-l+1)*(l+r)/2;
    }
    cout<<ans;
    return 0;
}