题意：

在这里插入图片描述

思路：

我们设 $s(x)=∑i=1nf(x)s(x)=\sum_{i=1}^nf(x)$ ，那么答案就是 $s (r) - s (l - 1)$ 。
容易发现，我们要求的 $f (x)$ 实际上就是 $x$ 的因子的个数，那么 $s(x)=∑i=1n∑d∣i1s(x)=\sum_{i=1}^n\sum_{d|i}1$ ，我们改为枚举因子，即 $s(x)=∑i=1n⌊ni⌋s(x)=\sum_{i=1}^n\left \lfloor \frac{n}{i} \right \rfloor$ ，这个可以整除分块 $O(n)O(\sqrt n )$ 求。

// Problem: P3935 Calculating
// Contest: Luogu
// URL: https://www.luogu.com.cn/problem/P3935
// Memory Limit: 128 MB
// Time Limit: 1000 ms
// 
// Powered by CP Editor (https://cpeditor.org)//#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4.1,sse4.2,avx,avx2,popcnt,tune=native")
//#pragma GCC optimize(2)
#include<cstdio>
#include<iostream>
#include<string>
#include<cstring>
#include<map>
#include<cmath>
#include<cctype>
#include<vector>
#include<set>
#include<queue>
#include<algorithm>
#include<sstream>
#include<ctime>
#include<cstdlib>
#include<random>
#include<cassert>
#define X first
#define Y second
#define L (u<<1)
#define R (u<<1|1)
#define pb push_back
#define mk make_pair
#define Mid ((tr[u].l+tr[u].r)>>1)
#define Len(u) (tr[u].r-tr[u].l+1)
#define random(a,b) ((a)+rand()%((b)-(a)+1))
#define db puts("---")
using namespace std;//void rd_cre() { freopen("d://dp//data.txt","w",stdout); srand(time(NULL)); }
//void rd_ac() { freopen("d://dp//data.txt","r",stdin); freopen("d://dp//AC.txt","w",stdout); }
//void rd_wa() { freopen("d://dp//data.txt","r",stdin); freopen("d://dp//WA.txt","w",stdout); }typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int,int> PII;const int N=1000010,mod=998244353,INF=0x3f3f3f3f;
const double eps=1e-6;LL l,r;LL get(LL n) {LL ans=0;for(LL l=1,r;l<=n;l=r+1) {r=n/(n/l);ans+=(r-l+1)%mod*((n/l)%mod)%mod; ans%=mod;}return ans;
} int main()
{
//	ios::sync_with_stdio(false);
//	cin.tie(0);LL ans=0;scanf("%lld%lld",&l,&r);ans+=get(r)-get(l-1); ans%=mod; ans+=mod; ans%=mod;printf("%lld\n",ans);return 0;
}
/**/