[洛谷3935]Calculating

题目链接：https://www.luogu.org/problemnew/show/P3935

首先显然有\(\sum\limits_{i=l}^rf(i)=\sum\limits_{i=1}^rf(i)-\sum\limits_{i=1}^{l-1}f(i)\)，于是问题转化为了如何求\(\sum\limits_{i=1}^nf(i)\)，即\(\sum\limits_{i=1}^n\sum\limits_{d|i}1\)，调整枚举顺序有\(\sum\limits_{d=1}^n\sum\limits_{i=1}^{\lfloor\frac{n}{d}\rfloor}1\)，即\(\sum\limits_{d=1}^n\lfloor\dfrac{n}{d}\rfloor\)，由于\(n\)很大，所以我们使用整除分块即可

/*program from Wolfycz*/
#include<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define inf 0x7f7f7f7f
using namespace std;
typedef long long ll;
typedef unsigned int ui;
typedef unsigned long long ull;
inline char gc(){
	static char buf[1000000],*p1=buf,*p2=buf;
	return p1==p2&&(p2=(p1=buf)+fread(buf,1,1000000,stdin),p1==p2)?EOF:*p1++;
}
inline int frd(){
	int x=0,f=1; char ch=gc();
	for (;ch<'0'||ch>'9';ch=gc())	if (ch=='-')	f=-1;
	for (;ch>='0'&&ch<='9';ch=gc())	x=(x<<3)+(x<<1)+ch-'0';
	return x*f;
}
inline int read(){
	int x=0,f=1; char ch=getchar();
	for (;ch<'0'||ch>'9';ch=getchar())	if (ch=='-')	f=-1;
	for (;ch>='0'&&ch<='9';ch=getchar())	x=(x<<3)+(x<<1)+ch-'0';
	return x*f;
}
inline void print(int x){
	if (x<0)	putchar('-'),x=-x;
	if (x>9)	print(x/10);
	putchar(x%10+'0');
}
const int p=998244353;
int solve(ll n){
	int res=0;
	for (ll i=1,pos;i<=n;i=pos+1){
		pos=n/(n/i); int len=(pos-i+1)%p;
		res=(res+1ll*len*(n/i)%p)%p;
	}
	return res;
}
int main(){
	ll l,r;
	scanf("%lld%lld",&l,&r);
	printf("%d\n",(solve(r)-solve(l-1)+p)%p);
	return 0;
}