Luogu 4240：毒瘤之神的考验

Posted 2021-01-29 cjoieryl

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Sol

分开考虑 (varphi(ij)) 中 (ij) 的质因子
那么
[varphi(ij)=frac{varphi(i)varphi(j)gcd(i,j)}{varphi(gcd(i,j))}]
直接莫比乌斯反演
设 (g(x,i)=sum_{j=1}^{x}varphi(ij))
那么
[ans=sum_{i=1}^{min(n,m)}g(lfloorfrac{n}{i} floor,i)g(lfloorfrac{m}{i} floor,i)sum_{d|i}mu(frac{i}{d})frac{d}{varphi(d)}]
后面的卷积可以直接筛
(Theta(Tn)) 当然不行了
设
(f(i)=sum_{d|i}mu(frac{i}{d})frac{d}{varphi(d)})
(s(i,j,k)) 表示 (sum_{p=1}^{k}g(i,p)g(j,p)f(p))
考虑到当 (ile sqrt{n}) 的时候 (s(i,j,k)) 中的 (i,j ge sqrt{n})
当 (ige sqrt{n}) 的时候 (s(i,j,k)) 中的 (i,j le sqrt{n})
所以可以预处理到 (s(80,80,k)) 对于小于 (n/80) 的直接暴力

# include <bits/stdc++.h>
using namespace std;
typedef long long ll;

const int mod(998244353);
const int maxn(1e5);
const int blk(80);

inline void Inc(int &x, int y) {
    if ((x += y) >= mod) x -= mod;
}

int pr[maxn + 100], phi[maxn + 100], inv[maxn + 100], mu[maxn + 100], tot, f[maxn + 100];
int test, n, m, ans;
vector <int> g[maxn + 100], s[blk + 1][blk + 1];
bitset <maxn + 100> ispr;
  
int main() {
    register int i, j, k, len;
    mu[1] = phi[1] = inv[1] = 1, ispr[1] = 1;
    for (i = 2; i <= maxn; ++i) inv[i] = (ll)(mod - mod / i) * inv[mod % i] % mod;
    for (i = 2; i <= maxn; ++i) {
        if (!ispr[i]) pr[++tot] = i, mu[i] = -1, phi[i] = i - 1;
        for (j = 1; j <= tot && i * pr[j] <= maxn; ++j) {
            ispr[i * pr[j]] = 1;
            if (i % pr[j]) mu[i * pr[j]] = -mu[i], phi[i * pr[j]] = phi[i] * (pr[j] - 1);
            else {
                mu[i * pr[j]] = 0;
                phi[i * pr[j]] = phi[i] * pr[j];
                break;
            }
        }
    }
    for (i = 1; i <= maxn; ++i) Inc(mu[i], mod);
    for (i = 1; i <= maxn; ++i)
        for (j = i; j <= maxn; j += i) Inc(f[j], (ll)i * inv[phi[i]] % mod * mu[j / i] % mod);
    for (i = 1; i <= maxn; ++i) {
        len = maxn / i, g[i].resize(len + 1);
        for (j = 1; j <= len; ++j) g[i][j] = (g[i][j - 1] + phi[i * j]) % mod;
    }
    for (i = 1; i <= blk; ++i)
        for (j = i; j <= blk; ++j) {
            len = maxn / j, s[i][j].resize(len + 1);
            for (k = 1; k <= len; ++k) s[i][j][k] = (s[i][j][k - 1] + (ll)f[k] * g[k][i] % mod * g[k][j] % mod) % mod;
        }
    for (scanf("%d", &test); test; --test) {
        ans = 0, scanf("%d%d", &n, &m);
        if (n > m) swap(n, m);
        len = min(n, m / blk);
        for (i = 1; i <= len; ++i) Inc(ans, (ll)g[i][n / i] * g[i][m / i] % mod * f[i] % mod);
        for (i = len + 1; i <= n; i = j + 1) {
            j = min(n / (n / i), m / (m / i));
            Inc(ans, (s[n / i][m / i][j] - s[n / i][m / i][i - 1] + mod) % mod);
        }
        printf("%d
", ans);
    }
    return 0;
}