牛客多校七 Dividing(数论分块)

数论分块：求n/1+n/2+n/3+…+n/k的和

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define ll long long
ll mod=1e9+7;
int main()
{
    ll n,k;
     
    scanf("%lld%lld",&n,&k);
    ll ans=(n+k-1)%mod;
    ll gx;
    for(ll i=2;i<=k;i=gx+1)
    {
        gx=n/i?min(k,n/(n/i)):k;
        ans+=(gx-i+1)%mod*(n/i)%mod;
        ans%=mod;//一定要模
    }
    n--;
    for(ll i=2;i<=k;i=gx+1)
    {
        gx=n/i?min(k,n/(n/i)):k;
        ans+=(gx-i+1)%mod*(n/i)%mod;
        ans%=mod;//一定要模
    }
    printf("%lld\n",ans);
}