题解：a%b=a-[a/b]*b;
ans=n*k-sigma([k/i]*i]);
[k/i]有sqrt(k)种取值
last=k/(k/last);
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
typedef long long Lint;
int n,m;
Lint ans;
int main(){
cin>>n>>m;
ans=1LL*n*m;
if(m>n){
int last;
for(int i=1;i<=n;i=last+1){
last=min(n,m/(m/i));
ans=ans-(last-i+1)*1LL*(last+i)*(m/i)/2;
}
}else{
int last;
for(int i=1;i<=m;i=last+1){
last=m/(m/i);
ans=ans-(last-i+1)*1LL*(last+i)*(m/i)/2;
}
}
cout<<ans<<endl;
return 0;
}