HDU6750 Function
题解
先将题意转化,我们得到
∑i=1n∑t∣it[gcd(t,it)=1]\displaystyle\sum_{i=1}^{n}\displaystyle\sum_{t|i}t[gcd(t,\frac{i}{t})=1] i=1∑nt∣i∑t[gcd(t,ti)=1]
化简这个式子即可
推导
∑i=1n∑t∣it[gcd(t,it)=1]=∑i=1n∑t∣it∑d∣t,dt∣iμ(d)=∑d=1nμ(d)∑t=1⌊n/d⌋td∑i=1⌊n/d2t⌋1T=d2t→=∑T=1n(∑i=1⌊n/T⌋1)∑d2∣Tμ(d)Td=∑T=1n⌊nT⌋∑d2∣Tμ(d)Td=∑d=1nμ(d)d∑T=1⌊n/d2⌋⌊n/d2T⌋TG(k)=∑i=1k⌊ki⌋i→=∑d=1nμ(d)dG(⌊nd2⌋)
\begin{aligned}
\displaystyle\sum_{i=1}^{n}\displaystyle\sum_{t|i}t[gcd(t,\frac{i}{t})=1]
& = \displaystyle\sum_{i=1}^{n}\displaystyle\sum_{t|i} t \displaystyle\sum_{d|t,dt|i}\mu(d) \\
& = \displaystyle\sum_{d = 1} ^ {n} \mu(d) \displaystyle\sum_{t = 1} ^ {\lfloor n / d \rfloor} td \displaystyle\sum_{i = 1} ^ {\lfloor n / d ^ 2 t \rfloor} 1 \\
T = d ^ 2 t \to & = \displaystyle\sum_{T = 1} ^ {n} \left( \displaystyle\sum_{i = 1} ^ {\lfloor n / T \rfloor} 1 \right) \displaystyle\sum_{d ^ 2 | T} \mu(d) \frac{T}{d} \\
& = \displaystyle\sum_{T = 1} ^ {n} \lfloor \frac{n}{T} \rfloor \displaystyle\sum_{d ^ 2 | T} \mu(d) \frac{T}{d} \\
& = \displaystyle\sum_{d = 1} ^ {\sqrt n} \mu(d) d \displaystyle\sum_{T = 1} ^ {\lfloor n / d ^ 2 \rfloor} \lfloor \frac{n / d ^ 2}{T} \rfloor T \\
G(k) = \displaystyle\sum_{i = 1} ^ {k} \lfloor \frac{k}{i} \rfloor i \to & = \displaystyle\sum_{d = 1} ^ {\sqrt n} \mu(d) d G(\lfloor \frac{n}{d ^ 2} \rfloor)
\end{aligned}
i=1∑nt∣i∑t[gcd(t,ti)=1]T=d2t→G(k)=i=1∑k⌊ik⌋i→=i=1∑nt∣i∑td∣t,dt∣i∑μ(d)=d=1∑nμ(d)t=1∑⌊n/d⌋tdi=1∑⌊n/d2t⌋1=T=1∑n⎝⎛i=1∑⌊n/T⌋1⎠⎞d2∣T∑μ(d)dT=T=1∑n⌊Tn⌋d2∣T∑μ(d)dT=d=1∑nμ(d)dT=1∑⌊n/d2⌋⌊Tn/d2⌋T=d=1∑nμ(d)dG(⌊d2n⌋)
这里G函数和原式子都可以用数论分块来做,总体复杂度O(nlog(n))O(\sqrt n log(n))O(nlog(n))
代码
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int MAX = 1e6 + 10;
const ll mod = 1e9 + 7;
const ll inv2 = 500000004;
int vis[MAX], prime[MAX], num, mu[MAX], f[MAX];
ll g[MAX];
void makeMobius(int siz) {
mu[1] = 1, num = 0;
for (int i = 2; i <= siz; i++) {
if (!vis[i]) prime[++num] = i, mu[i] = -1;
for (int j = 1; j <= num && i * prime[j] <= siz; j++) {
vis[i * prime[j]] = 1;
if (i % prime[j] == 0) {
mu[i * prime[j]] = 0;
break;
}
else mu[i * prime[j]] = mu[i] * mu[prime[j]];
}
}
for (int d = 1; d <= siz; d++)
for (int i = d; i <= siz; i += d)
g[i] = (g[i] + d) % mod;
for (int i = 1; i <= siz; i++) {
f[i] = (f[i - 1] + 1ll * mu[i] * i % mod) % mod;//求mu * i的前缀和
g[i] = (g[i] + g[i - 1]) % mod;
}
}
ll sum(ll x) {
x %= mod;
return x * (x + 1) % mod * inv2 % mod;
}
ll calc(ll n) {
if (n <= 1e6) return g[n];//没有提前筛这东西, T了一发
ll res = 0;
for (ll l = 1, r; l <= n; l = r + 1) {
r = n / (n / l);
res = (res + 1ll * (sum(r) - sum(l - 1) + mod) % mod * (n / l) % mod) % mod;
}
return res;
}
int main() {
makeMobius(1e6);
int T; scanf("%d", &T);
while (T--) {
ll n; scanf("%lld", &n);
ll ans = 0;
for (ll l = 1, r; l * l <= n; l = r + 1) {
r = l;
while (n / ((r + 1) * (r + 1)) == n / (r * r)) r++;//这里分块不太会,就这样写了
ans = (ans + 1ll * calc(n / (l * l)) * ((f[r] - f[l - 1] + mod) % mod) % mod) % mod;
}
printf("%lld\n", (ans + mod) % mod);
}
return 0;
}