数论分块——2020-camp-day1-C

对一个向下整除的式子快速求和

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
//typedef __int128 LL;
//typedef unsigned long long ull;
//#define F first
//#define S second
typedef long double ld;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
typedef pair<ld,ld> pdd;
const ld PI=acos(-1);
const ld eps=1e-9;
//unordered_map<int,int>mp;
#define ls (o<<1)
#define rs (o<<1|1)
#define pb push_back
//#define a(i,j) a[(i)*(m+2)+(j)]  //m是矩阵的列数
//pop_back()
const int seed=131;
const int M = 1e5+7;
const ll mod=998244353;
/*
int head[M],cnt;
void init(){cnt=0,memset(head,0,sizeof(head));}
struct EDGE{int to,nxt,val;}ee[M*2];
void add(int x,int y,int z){ee[++cnt].nxt=head[x],ee[cnt].to=y,ee[cnt].val=z,head[x]=cnt;}
*//*
ll cal(ll x)//x个点的满边图 
{
 
 return ; 
}*/
ll qpow(ll a,ll b)
{
 ll ans=1;
 while(b)
 {
  if(b&1)ans=ans*a%mod;
  a=a*a%mod;
  b=b/2;
 }
 return ans%mod;
}
int main()
{
 int t;
 cin>>t;
 while(t--)
 {
  ll n,l,r;
  scanf("%lld%lld%lld",&n,&l,&r);
  ll ans=0;
  for(ll i=l,gx;i<=r;i=gx+1)
  {
   gx=(n/i)?min(n/(n/i),r):r;
   ll x=n/gx;
   ll tp=((i*(((((1+x)*x%mod)*(n-x))%mod-((x*(x+1)%mod)*(n-x-1))%mod+mod)%mod)%mod+n*( ((x+1)*(n-x-1)%mod +mod-x*(n-x)%mod)%mod )%mod    +     gx*(  ((((1+x)*x%mod)*(n-x)%mod)+mod-(x*(x+1)%mod)*(n-x-1+2ll*mod)%mod )%mod )%mod+n*( ( (x+1)*(n-x-1+2ll*mod)%mod-x*(n-x)%mod +mod  )%mod )%mod  )*(gx-i+1)%mod)*qpow(2,mod-2)%mod;
   ans=(tp*qpow(2,mod-2)+ans)%mod;
  }
  printf("%lld
",ans);
 }
 return 0;
}