求$$\sum_{i=1}^{n} \gcd(\lfloor \sqrt[3]{i} \rfloor, i)$$
题解写的很清楚，自己重新推一推。
$$\sum_{i=1}^{n} \gcd(\lfloor \sqrt[3]{i} \rfloor, i)$$
$$=\sum_{a=1}^{\lfloor\sqrt[3]{n}\rfloor}\sum_{i=1}^{n}\gcd(a, i)[\sqrt[3]{i}=a]$$
$$=\sum_{a=1}^{\lfloor\sqrt[3]{n}\rfloor}\sum_{i=a^3}^{\min\{(a+1)^3-1,n\}}\gcd(a,i)$$
$$=\sum_{i=\lfloor \sqrt[3]{n} \rfloor ^3}^{n}\gcd(\sqrt[3]{n}, i)+\sum_{a=1}^{r}\sum_{i=a^3}^{(a+1)^3-1}\gcd(a,i)$$
其中 $r = \sqrt[3]{n}-1$
设 $f(n,a)=\sum_{i=1}^{n}\gcd(i, a)$
$$f(n,a)=\sum_{i=1}^{n}\gcd(i,a)$$
$$=\sum_{d|a}d\sum_{i=1}^{n}[\gcd(i,a)=d]$$
$$=\sum_{d|a}d\sum_{i=1}^{\lfloor \frac{n}{d} \rfloor}[\gcd(i, \frac{a}{d}) =1]$$
$$=\sum_{d|a}d\sum_{i=1}^{\lfloor \frac{n}{d} \rfloor}\sum_{p|\gcd(i,\frac{a}{d})}\mu(p)$$
$$=\sum_{d|a}d\sum_{p|\frac{a}{d}}\mu(p)\sum_{i=1}^{\lfloor \frac{n}{d} \rfloor}[p|i]$$
$$=\sum_{d|a}d\sum_{p|\frac{a}{d}}\mu(p)\lfloor \frac{n}{pd} \rfloor$$
$$=\sum_{T|a}\lfloor \frac{n}{T}\rfloor\sum_{p|T}\mu(p)\frac{T}{p}$$
$$=\sum_{T|a}\lfloor \frac{n}{T}\rfloor \varphi(T)$$
第一部分可以 $O(\sqrt{n})$ 解决。
将 $f(n, a)$ 带入第二部分得
$$\sum_{a=1}^r \sum_{i=a}^{(a+1)^3-1}\gcd(a,i)$$
$$=\sum_{a=1}^{r}\sum_{T|a}(\lfloor\frac{(a+1)^3-1}{T}\rfloor - \lfloor\frac{a^3-1}{T}\rfloor)\varphi(T)$$
$$=\sum_{T = 1}^{r}\varphi(T)\sum_{b=1}^{\lfloor\frac{r}{T}\rfloor}(\lfloor\frac{(bT+1)^3-1}{T} \rfloor-\lfloor\frac{(bT)^3-1}{T} \rfloor)$$
$$=\sum_{T=1}^{r}\varphi(T)\sum_{b=1}^{\lfloor\frac{r}{T}\rfloor}(\lfloor b^3T^2+3b^2T+3b\rfloor-\lfloor b^3T^2-\frac{1}{T}\rfloor)$$
$$=\sum_{T=1}^{r}\varphi(T)\sum_{b=1}^{\lfloor\frac{r}{T}\rfloor}(3b^2T+3b+1)$$
$$=\sum_{T=1}^{r}\varphi(T)(3T\sum_{b=1}^{\lfloor\frac{r}{T}\rfloor}b^2+3\sum_{b=1}^{\lfloor\frac{r}{T}\rfloor}b + \lfloor\frac{r}{T}\rfloor)$$
这部分 $O(r)$ 解决。
用太多int128会T。
#include <bits/stdc++.h> namespace IO { void read() {} template <typename T, typename... T2> inline void read(T &x, T2 &... oth) { T f = 1; x = 0; char ch = getchar(); while (!isdigit(ch)) { if (ch == '-') f = -1; ch = getchar(); } while (isdigit(ch)) { x = x * 10 + ch - 48; ch = getchar(); } x *= f; read(oth...); } } #define read IO::read #define print IO::print #define ll long long #define int128 __int128 const int MOD = 998244353; const int inv6 = 166374059, inv2 = 499122177; const int N = 1e7 + 7; int prime[N], phi[N], prin; bool vis[N]; void init() { phi[1] = 1; for (int i = 2; i < N; i++) { if (!vis[i]) { prime[++prin] = i; phi[i] = i - 1; } for (int j = 1; j <= prin && i * prime[j] < N; j++) { vis[i * prime[j]] = 1; if (i % prime[j] == 0) { phi[i * prime[j]] = prime[j] * phi[i]; break; } phi[i * prime[j]] = phi[i] * phi[prime[j]]; } } } int root3(int128 n) { int l = 1, r = 1e7 + 7; int ans = 1; while (l <= r) { int128 mid = (l + r) / 2; if (mid * mid * mid <= n) l = mid + 1, ans = mid; else r = mid - 1; } return ans; } template<class T> T gcd(T a, T b) { while (b) { a %= b; std::swap(a, b); } return a; } void M(int &a) { if (a < 0) a += MOD; if (a >= MOD) a -= MOD; } int f(int128 n, int a) { int ans = 0; for (int i = 1; 1LL * i * i <= a; i++) { if (a % i) continue; int128 temp = n / i * phi[i] % MOD; M(ans += temp); if (a == i * i) continue; int j = a / i; temp = n / j * phi[j] % MOD; M(ans += temp); } return ans; } int sum_squr(int n) { int ans = 1LL * n * (n + 1) % MOD * (2 * n + 1) % MOD * inv6 % MOD; return ans; } int sum(int n) { return 1LL * n * (n + 1) / 2 % MOD; } int main() { init(); int T; read(T); while (T--) { int128 n; read(n); if (n <= 7) { int ans = 0; for (int i = 1; i <= n; i++) ans += gcd(1, i); printf("%d\n", ans); continue; } int r = root3(n); int128 rr = (int128)r * r * r; int ans = f(n, r) - f(rr - 1, r); M(ans); r--; for (int i = 1; i <= r; i++) { int y = r / i; int temp = 3LL * i * sum_squr(y) % MOD; M(temp += 3LL * sum(y) % MOD); M(temp += y); temp = 1LL * temp * phi[i] % MOD; M(ans += temp); } printf("%d\n", ans); } return 0; }