tsy’s number

推式子

$$\begin{gathered} \sum _{i=1}^n\sum _{j=1}^n\sum _{k=1}^n\frac{?(i)?(j^2)?(k^3)}{?(i)?(j)?(k)}?(gcd(i,j,k)) \\ 我們假定j=p_1^{k_1}{p}_2^{k_2}{p}_3^{p_3}\dots \dots p_n^{k_n}，有?(j)=p_1^{k_1?1}(p_1?1)p_2^{k_2?1}(p_2?1)p_3^{k_3?1}(p_3?1)\dots \dots p_n^{k_n?1}(p_n?1)， \\ ?(j^2)=p_1^{2k_1?1}(p_1?1)p_2^{2k_2?1}(p_2?1)p_3^{2k_3?1}(p_3?1)\dots \dots p_n^{2k_n?1}(p_n?1) \\ \frac{?(j^2)}{?(j)}=p_1^{k_1}{p}_2^{k_2}{p}_3^{p_3}\dots \dots p_n^{k_n}=j \\ 同樣的，我們不難推出\frac{?(k^3)}{?(k)}=k^2 \\ 對上式化簡： \\ =\sum _{i=1}^n\sum _{j=1}^n\sum _{k=1}^{n}j{k}^2?(gcd(i,j,k)) \\ =\sum _{d=1}^n\sum _{i=1}^n\sum _{j=1}^n\sum _{k=1}^{n}j{k}^2?(d)(gcd(i,j,k)=d) \\ =\sum _{d=1}^n{d}^3?(d)\sum _{i=1}^{\frac{n}{d}}\sum _{j=1}^{\frac{n}{d}}\sum _{k=1}^{\frac{n}{d}}j{k}^2(gcd(i,j,k)=1) \\ =\sum _{d=1}^n{d}^3?(d)\sum _{t=1}^{\frac{n}{d}}\mu (t)t^3\sum _{i=1}^{\frac{n}{dt}}\sum _{j=1}^{\frac{n}{dt}}\sum _{k=1}^{\frac{n}{dt}}j{k}^2 \\ 設f(n)=\sum _{i=1}^n\sum _{j=1}^n\sum _{k=1}^{n}j{k}^2，T=dt，再次對上式化簡： \\ =\sum _{T=1}^{n}f(\frac{n}{T})T^3\sum _{d\mid T}?(d)\mu (\frac{T}{d}) \\ 我們假設g(n)=n^3\sum _{d\mid n}?(d)\mu (\frac{n}{d}) \end{gathered}$$

$$\begin{gathered} 顯然n^3是較好解決的，所以我們考慮后面的式子\sum _{d\mid n}?(d)\mu (\frac{n}{d}) \\ 寫成迪利克雷卷積的形式g(n)=(??\mu )(n)，顯然g(n)是一個積性函數 \\ n=1,g(1)=?(1)\mu (1)=1 \\ n=p,g(p)=?(1)\mu (p)+?(p)\mu (1)=1?(?1)+(p?1)?1=p?2 \\ n=p^{k}d，由于p^k與d互質，所以有g(n)=g(p^k)g(d) \\ 所以我們重點考慮如何求g(p^k) \\ g(p^k)=\sum _{i=0}^k?(p^i)\mu (p^{k?i}) \\ =\sum _{i=0}^k?(p^{k?i})\mu (p^i) \\ 顯然對于i\ge 2時\mu (p^i)=0,所以 \\ g(p^k)=\sum _{i=0}^1?(p^{k?i})\mu (p^i)=?(p^k)??(p^{k?1}) \\ 這里預處理就結束了，接下來就能快樂的數論分塊了。 \end{gathered}$$

代碼

/*Author : lifehappy */ #pragma GCC optimize(2) #pragma GCC optimize(3) #include <bits/stdc++.h>#define mp make_pair #define pb push_back #define endl '\n' #define mid (l + r >> 1) #define lson rt << 1, l, mid #define rson rt << 1 | 1, mid + 1, r #define ls rt << 1 #define rs rt << 1 | 1using namespace std;typedef long long ll; typedef unsigned long long ull; typedef pair<int, int> pii;const double pi = acos(-1.0); const double eps = 1e-7; const int inf = 0x3f3f3f3f;inline ll read() {ll f = 1, x = 0;char c = getchar();while(c < '0' || c > '9') {if(c == '-') f = -1;c = getchar();}while(c >= '0' && c <= '9') {x = (x << 1) + (x << 3) + (c ^ 48);c = getchar();}return f * x; }const int N = 1e7 + 10, mod = 1 << 30, inv3 = 715827883;int prime[N], cnt;ll f[N], g[N], n;bool st[N];ll quick_pow(ll a, int n) {ll ans = 1;while(n) {if(n & 1) ans = ans * a % mod;a = a * a % mod;n >>= 1;}return ans; }void init() {g[1] = 1;for(int i = 2; i < N; i++) {if(!st[i]) {prime[cnt++] = i;g[i] = i - 2;}for(int j = 0; j < cnt && i * prime[j] < N; j++) {st[i * prime[j]] = 1;if(i % prime[j] == 0) {int num = 1, temp = i;while(temp % prime[j] == 0) {temp /= prime[j], num++;}g[i * prime[j]] = quick_pow(prime[j], num - 2) * ((1ll * prime[j] * prime[j] % mod - 2 * prime[j] % mod + 1 + mod) % mod) % mod * g[temp] % mod;break;}g[i * prime[j]] = g[i] * g[prime[j]];}}for(int i = 1; i < N; i++) {f[i] = ((1ll * i * (1 + i) / 2 % mod) * i % mod) * ((1ll * (i + 1) * i / 2) % mod * (2 * i + 1) % mod * inv3 % mod) % mod;g[i] = (1ll * i * i % mod * i % mod * g[i] % mod + g[i - 1]) % mod;} }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 = read();while(T--) {n = read();ll ans = 0;for(ll l = 1, r; l <= n; l = r + 1) {r = n / (n / l);ans = (ans + 1ll * f[n / l] * ((g[r] - g[l - 1] + mod) % mod) % mod) % mod;}printf("%lld\n", ans);}return 0; }

總結

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