Luogu3768: 简单的数学题

题面

传送门

Sol

运用提出gcd等莫比乌斯反演的推导技巧得到

\[ans=\sum_{d=1}^{n}d^3\sum_{i=1}^{\lfloor\frac{n}{d}\rfloor}\mu(i)*i^2*S(\lfloor\frac{n}{d*i}\rfloor)^2 \]

其中\(S(n)=\sum_{i=1}^{n}i=\frac{n(n+1)}{2}\)
用\(代替d*i\)
\(ans=\sum_{i=1}^{n}S(\lfloor\frac{n}{i}\rfloor)^2 * i^2 \sum_{d|i}d*\mu(\frac{i}{d})\)
然后写线性筛就可以拿到优秀的暴力分

优化？

n这么大，想一想杜教筛
筛的时候发现\(\sum_{d|i}d*\mu(\frac{i}{d})\)和\(\varphi(d)\)的筛法一模一样
好的，那它们两个就相等
所以就是\(\sum_{i=1}^{n}S(\lfloor\frac{n}{i}\rfloor)^2 * i^2 * \varphi(i)\)
\(S(\lfloor\frac{n}{i}\rfloor)^2就分块一下\)后面的杜教筛

# include <bits/stdc++.h>
# define RG register
# define IL inline
# define Fill(a, b) memset(a, b, sizeof(a))
using namespace std;
typedef long long ll;
const int _(5e6 + 1);

IL ll Read(){
    RG ll x = 0, z = 1; RG char c = getchar();
    for(; c < '0' || c > '9'; c = getchar()) z = c == '-' ? -1 : 1;
    for(; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
    return x * z;
}

int prime[_], num, Zsy, phi[_], inv2, inv6, N;
map <ll, int> Phi;
bool isprime[_];

IL void Prepare(){
	isprime[1] = 1; phi[1] = 1;
	for(RG int i = 2; i < N; ++i){
		if(!isprime[i]){  prime[++num] = i; phi[i] = (i - 1);  }
		for(RG int j = 1; j <= num && i * prime[j] < N; ++j){
			isprime[i * prime[j]] = 1;
			if(i % prime[j]) phi[i * prime[j]] = 1LL * phi[i] * (prime[j] - 1) % Zsy;
			else{  phi[i * prime[j]] = 1LL * phi[i] * prime[j] % Zsy; break;  }
		}
	}
	for(RG int i = 2; i < N; ++i) phi[i] = 1LL * phi[i] * i % Zsy * i % Zsy, (phi[i] += phi[i - 1]) %= Zsy;
}

IL ll Pow(RG ll x, RG ll y){
	RG ll ret = 1;
	for(; y; x = x * x % Zsy, y >>= 1) if(y & 1) ret = ret * x % Zsy;
	return ret;
}

IL ll Ssqr(RG ll x){  return x % Zsy * (x + 1) % Zsy * (2 * x % Zsy + 1) % Zsy * inv6 % Zsy;  }

IL ll Sumphi(RG ll n){
	if(n < N) return phi[n];
	if(Phi[n]) return Phi[n];
	RG ll ans = n % Zsy * (n + 1) % Zsy * inv2 % Zsy; ans = ans * ans % Zsy;
	for(RG ll i = 2, j; i <= n; i = j + 1){
		j = n / (n / i);
		ans -= (Ssqr(j) - Ssqr(i - 1)) * Sumphi(n / i) % Zsy;
		ans = (ans + Zsy) % Zsy;
	}
	return Phi[n] = ans;
}

int main(RG int argc, RG char* argv[]){
	Zsy = Read(); RG ll n = Read(); N = min(1LL * _, n + 1); Prepare();
	inv2 = Pow(2, Zsy - 2); inv6 = Pow(6, Zsy - 2);
	RG ll ans = 0, s;
	for(RG ll i = 1, j; i <= n; i = j + 1){
		j = n / (n / i);
		s = (n / i) % Zsy * (n / i + 1) % Zsy * inv2 % Zsy;
		s = s * s % Zsy;
		ans += s * (Sumphi(j) - Sumphi(i - 1)) % Zsy;
		ans = (ans + Zsy) % Zsy;
	}
	printf("%lld\n", ans);
    return 0;
}