ICPC2019银川区域赛F. Function!

链接

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题解

$ans = \sum_{a=2}^n \left( a \sum_{b=a}^n \lfloor \log_ab \rfloor \lceil \log_ba \rceil \right) \\ = \sum_{a=2}^n \left( a \sum_{b=a}^n \lfloor \log_ab \rfloor\right) \\$

当 $a \le \sqrt n$时使用暴力

当 $a > \sqrt n$时，因为 $n < a^2$，所以 $\log_an<2$，所以 $1\le \log_ab \le log_an < 2$，即 $\lfloor \log_ab \rfloor \equiv 1$，此时 $a \sum_{b=a}^n \lfloor \log_ab \rfloor = a(n-a+1)$

代码

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#define iinf 0x3f3f3f3f
#define linf (1ll<<60)
#define eps 1e-8
#define maxn 1000010
#define maxe 1000010
#define cl(x) memset(x,0,sizeof(x))
#define rep(i,a,b) for(i=a;i<=b;i++)
#define drep(i,a,b) for(i=a;i>=b;i--)
#define em(x) emplace(x)
#define emb(x) emplace_back(x)
#define emf(x) emplace_front(x)
#define fi first
#define se second
#define de(x) cerr<<#x<<" = "<<x<<endl
using namespace std;
using namespace __gnu_pbds;
typedef long long ll;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
ll read(ll x=0)
{
    ll c, f(1);
    for(c=getchar();!isdigit(c);c=getchar())if(c=='-')f=-f;
    for(;isdigit(c);c=getchar())x=x*10+c-0x30;
    return f*x;
}
#define mod 998244353ll
ll inv[maxn];
ll S1(ll n)
{
    n%=mod;
    return n*(n+1)/2%mod;
}
ll S2(ll n)
{
    n%=mod;
    return n*(n+1) %mod * (2*n+1) %mod * inv[6] %mod;
}
int main()
{
    ll a, b, n=read(), ans=0;
    inv[1]=1;for(int i=2;i<maxn;i++)inv[i]=inv[mod%i]*(mod-mod/i)%mod;
    for(a=2;a*a<=n;a++)
    {
        ll t=1;
        for(b=a;b<=n;b*=a,t++)
        {
            ans += (min(b*a-1,n)-b+1) * a * t;
            ans %= mod;
        }
    }
    ans += (n+1)%mod*(S1(n)-S1(a-1))%mod - ( S2(n) - S2(a-1) )%mod;
    printf("%lld",(ans%mod+mod)%mod);
    return 0;
}