题意
$定义f_n(k)=\sum_{l_1=1}^n\sum_{l_2=1}^n…\sum_{l_k=1}^n[gcd(l_1,l_2,…,l_k)]^2$
$$求解\sum_{i=2}^kf_{n(i)}$$
思路
$$f_n(k)=\sum_{l_1=1}^n\sum_{l_2=1}^n…\sum_{l_k=1}^n[gcd(l_1,l_2,…,l_k)]^2$$
$$=\sum_{t=1}^n\sum_{l_1=1}^n\sum_{l_2=1}^n…\sum_{l_k=1}^n[gcd(l_1,l_2,…,l_k)=t]*t^2$$
$$=\sum_{t=1}^{n}t^2\sum_{l_1=1}^{\left \lfloor \frac{n}{t} \right \rfloor }\sum_{l_2=1}^{\left \lfloor \frac{n}{t} \right \rfloor }…\sum_{l_k=1}^{\left \lfloor \frac{n}{t} \right \rfloor }[gcd(l_1,l_2,…,l_k)=1]$$
$$=\sum_{t=1}^{n}t^2\sum_{l_1=1}^{\left \lfloor \frac{n}{t} \right \rfloor }\sum_{l_2=1}^{\left \lfloor \frac{n}{t} \right \rfloor }…\sum_{l_k=1}^{\left \lfloor \frac{n}{t} \right \rfloor }\sum_{dl_1\;…dl_k}\mu(d)$$
$$=\sum_{t=1}^nt^2\sum_{d=1}^{\left \lfloor \frac{n}{t} \right \rfloor}\mu(d)*\left \lfloor \frac{n}{dt} \right \rfloor^k$$
$令T=dt，则：$
$$f_n(k)=\sum_{T=1}^n\sum_{dT}\mu(d)\left \lfloor \frac{T}{d} \right \rfloor^2\left \lfloor \frac{n}{T} \right \rfloor^k$$
$$\sum_{i=2}^kf_n(i)=\sum_{i=2}^k\sum_{T=1}^n\sum_{dT}\mu(d)\left \lfloor \frac{T}{d} \right \rfloor^2\left \lfloor \frac{n}{T} \right \rfloor^i$$
$$\sum_{T=1}^n\sum_{dT}\mu(d)\left \lfloor \frac{T}{d} \right \rfloor^2\sum_{i=2}^k\left \lfloor \frac{n}{T} \right \rfloor^i$$
$令h(x)=\sum_{dx}\mu(d)\left \lfloor \frac{x}{d} \right \rfloor^2，设S(n)=\sum_{i=1}^nh(i)，用杜教筛求出h的前缀和。$
$根据杜教筛求前缀和公式：$
$$S(n)=\frac{\sum_{i=1}^n(f*g)i-\sum_{i=2}^ng[i]S(\left \lfloor \frac{n}{i} \right \rfloor)}{g[1]}$$
$为了简便公式，设g=I,则(h*g)i=(I*\mu *id^2)i=i^2$
$则$ $$S(n)=\frac{n(n+1)(2n+1)}{6}-\sum_{i=2}^nS(\left \lfloor \frac{n}{i} \right \rfloor)$$
$对于\sum_{i=2}^k\left \lfloor \frac{n}{T} \right \rfloor^i，利用等比公式，p=\left \lfloor \frac{n}{T} \right \rfloor.$
$$\sum_{i=2}^k\left \lfloor \frac{n}{T} \right \rfloor^i=\frac{(\frac{n}{T})^2*[1-(\frac{n}{T})^{k-1}]}{1-\frac{n}{T}}$$
$注意\frac{n}{T}=1的情况，答案为k-1。$
$因为k很大，所以可以用欧拉降幂预先处理再作等比，再配合杜教筛求前缀和。$
$$\sum_{i=2}^nf_n(i)=\sum_{T=1}^nh(i)*\frac{(\frac{n}{T})^2*[1-(\frac{n}{T})^{k-1}]}{1-\frac{n}{T}}$$
$总时间复杂度为O(T(n^{\frac{2}{3}}+\sqrt n\log n))$
Code(MS)
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| ##include "bits/stdc++.h"
using namespace std;
typedef long long ll;
##define endl "\n"
const ll mod = 1e9 + 7;
const int N = 1e5 + 10;
bool is_prime[N];
int prime[N], cnt;
ll h[N], s[N];
ll inv6, k1, k2;
void sieve() {
h[1] = 1;
is_prime[0] = is_prime[1] = true;
for(int i = 2;i < N; i++) {
if(!is_prime[i]) {
prime[++cnt] = i;
h[i] = 1ll * i * i % mod - 1;
}
for(int j = 1;j <= cnt && i * prime[j] < N; j++) {
is_prime[i * prime[j]] = true;
if(i % prime[j] == 0) {
h[i * prime[j]] = h[i] * prime[j] % mod * prime[j] % mod;
break;
}
else {
h[i * prime[j]] = h[i] * h[prime[j]] % mod;
}
}
}
for(int i = 1;i < N; i++) s[i] = (s[i - 1] + h[i]) % mod;
}
ll quick_pow(ll a, ll b) {
ll ans = 1;
while(b) {
if(b & 1) ans = ans * a % mod;
a = a * a % mod;
b >>= 1;
}
return ans % mod;
}
map<ll, ll> mp;
ll S(ll x) {
if(x < N) return s[x];
if(mp[x]) return mp[x];
ll ans = x * (x + 1) % mod * (2 * x + 1) % mod * inv6 % mod;
for(int l = 2, r;l <= x; l = r + 1) {
r = min(x / (x / l), x);
ans = (ans - 1ll * (r - l + 1) * S(x / l) % mod) % mod;
}
return mp[x] = (ans % mod + mod) % mod;
}
ll Sum(ll p) {
if(p == 1)
return k2 - 1;
else
return p * p % mod * (quick_pow(p, k1 - 1) - 1) % mod * quick_pow(p - 1, mod - 2) % mod;
}
void init() {
sieve();
inv6 = quick_pow(6, mod - 2);
}
void solve() {
init();
int _; cin >> _;
while(_--) {
k1 = k2 = 0;
int n; cin >> n;
string K; cin >> K;
for(int i = 0;i < K.length(); i++) {
k1 = (k1 * 10 + K[i] - '0') % (mod - 1);
k2 = (k2 * 10 + K[i] - '0') % mod;
}
k1 += mod - 1;
ll ans = 0;
ll pre = 0, now = 0;
for(int l = 1, r;l <= n; l = r + 1) {
r = min(n / (n / l), n);
now = S(r);
ans = (ans + (now - pre + mod) % mod * Sum(n / l) % mod) % mod;
pre = now;
}
cout << (ans % mod + mod) % mod << endl;
}
}
signed main() {
solve();
}
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