[EOJ] # 3349

　　A very interesting question.

　　Before solving the problem, you need to know:

　　　　$*$　$ \sum _{i=1}^{i=n} (\phi (i)* \lfloor \frac n i \rfloor)= \frac {n*(n+1)} 2$

　　　　$*$　$ \phi (i)= n*(\prod \frac 1 {p_i})$

//#pragma GCC optimize(2)
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll mod=998244353; 
ll n,k,tot,x,sum,ANS,F;
bool b[1000010];
int res[1000100],f[1000100][41],p[1000010];
inline void getp(ll n){
    for(ll i=2;i<=sqrt(n);i++){
        if(b[i])continue;
        p[++tot]=i;
        for(int j=2*i;j<=sqrt(n);j+=i)b[j]=1;
    }
} 
inline void euler()
{
    res[1]=1;
    for(int i=2;i<k;i++)
    {
        if(!res[i])
        {
            for(int j=i;j<k;j+=i)
            {
                if(!res[j])res[j]=j;
                res[j]=res[j]/i*(i-1);
            }
        }
    }
}
int main()
{
//    freopen("mogic.in","r",stdin);
//    freopen("mogic.out","w",stdout);
    cin>>n>>k;getp(n);euler();
    x=(n%mod)*((n+1)%mod)/2;
    for(ll i=1;i<k;i++)sum=(sum+res[i]*(n/i)%mod)%mod;
    for(ll i=1;i<=tot;i++){
        ll sq=ceil((double)(n-k+1)/(double)p[i])*p[i];
        for(ll j=sq;j<=n;j+=p[i])
        f[j-n+k][++f[j-n+k][0]]=i;
    }
    for(ll i=1;i<=k;i++){
        F=n-k+i;
        ll pr=n-k+i;
        for(int j=1;j<=f[i][0];j++){
            F=F/p[f[i][j]]*(p[f[i][j]]-1);
            while(!(pr%p[f[i][j]]))pr/=p[f[i][j]];
        }
        if(F==n-k+i)F--;
        else if(pr!=1)F=F/pr*(pr-1);
        sum=(sum+F)%mod;
    }
    ANS=(x-sum+mod)%mod;cout<<ANS;
    return 0;
}

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