直接for（int i=x;i<=n;i+=x）add(i,y)；最坏是o(n^m)=1e10，会爆掉
所以需要用个lazy[]进行分块统计小于sqrt(n)复杂度较大的那一部分y，
最后在算range_sum(x,y)的时候加上[x,y]中lazy[]数组的贡献

#pragma GCC optimize(2)
#include <bits/stdc++.h>
#define endl '\n'
#define ll long long 
using namespace std;
ll tree[100050],lazy[100050],n,m,ans;
ll lowbit(ll x)
{
    return x&-x;    
} 
void add(ll i,ll x)
{
    while(i<=n)
    {
        tree[i]+=x;
        i+=lowbit(i);
    }
}
ll sum(ll i)
{
    ll res=0;
    while(i>0)
    {
        res+=tree[i];
        i-=lowbit(i);
    }
    return res;    
}
ll range_sum(ll l,ll r)
{
    return sum(r)-sum(l-1);
}
int main()
{
    ios::sync_with_stdio(0);cin.tie(0), cout.tie(0);
      cin>>n>>m;
      while(m--)
      {
          ll f,x,y;
          cin>>f>>x>>y;
          if(f==1){
              if(x<=sqrt(n)){
                lazy[x]+=y;
            }
            else{
                for(ll i=x;i<=n;i+=x)
                    add(i,y);
            }  
        }
        else if(f==2) {
            ans=range_sum(x,y);
              for(ll i=1;i<=sqrt(n);i++)
              ans+=(y/i-(x-1)/i)*lazy[i];
              cout<<ans<<endl;     
        }
    }
    return 0;
}