Problem C. Calculate

$$\begin{gathered} ?=???=??\mu ?I \\ ?(n)=\sum _{d\mid n}(??\mu )(d) \\ 設g(n)=\sum _{d\mid n}(??\mu )(d) \\ \sum _{i=1}^A\sum _{j=1}^B\sum _{k=1}^C?(gcd(i,j^2,k^3)) \\ \sum _{i=1}^A\sum _{j=1}^B\sum _{k=1}^C\sum _{d\mid i,d\mid j^2,d\mid k^3}(??\mu )(d) \\ \sum _{d=1}^A(??\mu )(d)\sum _{i=1}^A\sum _{j=1}^B\sum _{k=1}^C[d\mid i,d\mid j^2,d\mid k^3] \end{gathered}$$
設$x=\prod p_i^{k_i}$，唯一分解，如果$x\mid y^t$，則有$\prod p_i^{?\frac{k_i}{t}?}\mid y$，設$f_t(n)=\prod p_j^{\frac{k_j}{t}}$。

上式可得：
$$\sum _{d=1}^A(??\mu )(d)?\frac{d}{f_1(d)}??\frac{d}{f_2(d)}??\frac{d}{f_3(d)}?$$

對于$h=(??\mu )(d),f_1,f_2,f_3$都具有積性，所以可以線性篩預處理，最后$O(A)$統計答案。

#include <bits/stdc++.h>using namespace std;typedef unsigned uint;const int N = 1e7 + 10;uint prime[N], h[N], f1[N], f2[N], f3[N], num[N], primes[N], cnt;bool st[N];void init() {h[1] = f1[1] = f2[1] = f3[1] = primes[1] = 1;for (int i = 2; i < N; i++) {if (!st[i]) {prime[++cnt] = i;h[i] = i - 2;f1[i] = i;f2[i] = i;f3[i] = i;primes[i] = i;num[i] = 1;}for (int j = 1; j <= cnt && 1ll * i * prime[j] < N; j++) {st[i * prime[j]] = 1;if (i % prime[j] == 0) {primes[i * prime[j]] = primes[i] * prime[j];num[i * prime[j]] = num[i] + 1;f1[i * prime[j]] = f1[i] * prime[j];f2[i * prime[j]] = f2[i / prime[j]] * prime[j];if (i / prime[j] % prime[j] == 0) {f3[i * prime[j]] = f3[i / prime[j] / prime[j]] * prime[j];}else {f3[i * prime[j]] = f3[i];}if(i * prime[j] == primes[i * prime[j]]) {h[i * prime[j]] = i * prime[j] - 2 * i + i / prime[j];}else {h[i * prime[j]] = h[i / primes[i]] * h[primes[i * prime[j]]];}break;}h[i * prime[j]] = h[i] * h[prime[j]];f1[i * prime[j]] = f1[i] * f1[prime[j]];f2[i * prime[j]] = f2[i] * f2[prime[j]];f3[i * prime[j]] = f3[i] * f3[prime[j]];primes[i * prime[j]] = prime[j];num[i * prime[j]] = 1;}} }int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);init();int T;scanf("%d", &T);while (T--) {uint A, B, C, ans = 0;scanf("%u %u %u", &A, &B, &C);for (int d = 1; d <= (int)A; d++) {ans += h[d] * (A / f1[d]) * (B / f2[d]) * (C / f3[d]);}printf("%u\n", ans & ((1 << 30) - 1));}return 0; }

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