[洛谷 P6055] [RC-02] GCD (莫比乌斯反演 杜教筛)

题意

求

$$\sum _{i=1}^n\sum _{j=1}^n\sum _{p=1}^{\lfloor \frac{n}{j}\rfloor }\sum _{q=1}^{\lfloor \frac{n}{j}\rfloor }[\gcd (i,j)=1][\gcd (p,q)=1]$$

对 $998244353$ 取模

分析：

常规套用莫比乌斯反演式子会很麻烦，所以这里反向把上界拿下来

$$\sum _{i=1}^n\sum _{j=1}^n\sum _{p=1}^n\sum _{q=1}^n[\gcd (i,j)=1][\gcd (p,q)=j]$$

那么就是

$$\sum _{i=1}^n\sum _{j=1}^n\sum _{k=1}^n[\gcd (i,j,k)=1]$$

再来莫比乌斯反演

$$\sum _{i=1}^n\sum _{j=1}^n\sum _{k=1}^n\sum _{d\mid \gcd (i,j,k)}\mu (d)$$

化简一下

$$\sum _{d=1}^n\mu (d)\lfloor \frac{n}{d}{\rfloor }^3$$

就可以用杜教筛了

代码：

#include 
#define int long long
using namespace std;
const int N = 2e6 + 5, mod = 998244353;
int n, mobius[N], primes[N], cnt, sum[N], res;
bool st[N];
unordered_map<int, int> mp;
void get_mobius(int n) {
    mobius[1] = 1;
    for (int i = 2; i <= n; i ++) {
        if (!st[i]) {
            primes[cnt ++] = i;
            mobius[i] = -1;
        }
        for (int j = 0; primes[j] * i <= n; j ++) {
            int t = primes[j] * i;
            st[t] = 1;
            if (i % primes[j] == 0) {
                mobius[t] = 0;
                break;
            }
            mobius[t] = -mobius[i];
        }
    }
    for (int i = 1; i <= n; i ++) sum[i] = (sum[i - 1] + mobius[i] + mod) % mod;
}
int Sum(int n) {
    if (n < N) return sum[n];
    if (mp[n]) return mp[n];
    int res = 1;
    for (int l = 2, r; l <= n; l = r + 1) {
        r = n / (n / l);
        res = (res - Sum(n / l) * (r - l + 1) % mod + mod) % mod;
    }
    return mp[n] = res;
}
signed main() {
    get_mobius(N - 1);
    cin >> n;
    for (int l = 1, r; l <= n; l = r + 1) {
        r = n / (n / l);
        res = (res + (Sum(r) - Sum(l - 1)) * (n / l) % mod * (n / l) % mod * (n / l) % mod + mod) % mod;
    }
    cout << res << endl;
}