题解 P6055 【[RC-02] GCD】

[ans=sum_{i=1}^Nsum_{j=1}^Nsum_{p=1}^{leftlfloordfrac{N}{j} ight floor}sum_{q=1}^{leftlfloordfrac{N}{j} ight floor}[gcd(i,j)=1][gcd(p,q)=1] ]

方法1：

[egin{aligned} ans & = sum_{i=1}^Nsum_{j=1}^Nsum_{p=1}^{leftlfloordfrac{N}{j} ight floor}sum_{q=1}^{leftlfloordfrac{N}{j} ight floor}[gcd(i,j)=1][gcd(p,q)=1] \ & = sum_{i=1}^{N}sum_{j=1}^Nsum_{p=1}^{N}sum_{q=1}^N[gcd(i,j)=1][gcd(p,q)=j] \ & = sum_{i=1}^Nsum_{p=1}^{N}sum_{q=1}^N[gcd(i,p,q)=1] \ & = sum_{d=1}^Nmu(d)leftlfloordfrac{N}{d} ight floor^3 end{aligned} ]

方法2:

[egin{aligned} ans & =sum_{i=1}^{N}sum_{j=1}^N(sum_{d|gcd(i,j)}mu(d))(sum_{p=1}^{leftlfloordfrac{N}{j} ight floor}sum_{q=1}^{leftlfloordfrac{N}{j} ight floor}sum_{t|gcd(p,q)}mu(t)) \ & =sum_{i=1}^{N}sum_{j=1}^N(sum_{d|gcd(i,j)}mu(d))(sum_{t=1}^{leftlfloordfrac{N}{j} ight floor}mu(t)leftlfloordfrac{N}{jt} ight floor^2) \ & =sum_{d=1}^Nmu(d)leftlfloordfrac{N}{d} ight floorsum_{j=1}^{leftlfloordfrac{N}{d} ight floor}1cdotsum_{t=1}^{leftlfloordfrac{N}{d} ight floor}leftlfloordfrac{N}{dt} ight floor^2sum_{g|t}mu(t) \ & =sum_{d=1}^Nmu(d)leftlfloordfrac{N}{d} ight floorsum_{j=1}^{leftlfloordfrac{N}{d} ight floor}1cdotsum_{t=1}^{leftlfloordfrac{N}{d} ight floor}leftlfloordfrac{N}{dt} ight floor^2[t=1] \ & =sum_{d=1}^Nmu(d)leftlfloordfrac{N}{d} ight floorsum_{j=1}^{leftlfloordfrac{N}{d} ight floor}1cdotleftlfloordfrac{N}{d} ight floor^2 \ & =sum_{d=1}^Nmu(d)leftlfloordfrac{N}{d} ight floor^3 end{aligned} ]

然后数论分块+杜教筛即可

#include<cstdio>
#include<map>
#define Starseven main
#define ll long long
const int N = 8e6;
const ll mod = 998244353;
int prime[N + 20], num;
ll mu[N + 20];
bool vis[N + 20];
std::map<ll,ll> mp;

void Init() {
	mu[1] = 1;
	for (int i = 2; i <= N; i++) {
		if(!vis[i]) {
			prime[++num] = i;
			mu[i] = -1ll;
		}
		for (int j = 1; j <= num && prime[j] * i <= N; j++) {
			int x = prime[j] * i;
			vis[x] = true;
			if(i % prime[j] == 0) {
				mu[x] = 0;
				break;
			}
			mu[x] = -mu[i];
		}
	}
	for (int i = 1; i <= N; i++) {
		mu[i] += mu[i - 1];
	} 
	return ;
}

ll Get_sum(ll x) {
	if(x <= N) return mu[x];
	if(mp[x]) return mp[x];
	ll re = 1;
	for (ll l = 2, r; l <= x; l = r + 1) {
		r = x / (x / l);
		re = (re - (r - l + 1) * Get_sum(x / l)) % mod;
		re = (re + mod) % mod;
	}
	mp[x] = re;
	return re;
}

int Starseven(void) {
	ll n;
	read(n);
	Init();
	ll ans = 0;
	for (ll l = 1, r; l <= n; l = r + 1) {
		ll hack = n / l;
		r = n / (n / l);
		hack = hack * hack % mod * hack % mod;
		ans = (ans + (Get_sum(r) - Get_sum(l - 1)) % mod * hack % mod) % mod;
		ans = (ans + mod) % mod;
	}
	write(ans);
	puts(""); 
	return 0;	
}