K Sum

推式子

$$\begin{gathered} F_n(k)=\sum _{l_1=1}^n\sum _{l_2=1}^n\cdots \sum _{l_k=1}^n(gcd(l_1,l_2,\dots ,l_k){)}^2 \\ =\sum _{d=1}^n{d}^2\sum _{l_1=1}^{\frac{n}{d}}\sum _{l_2=1}^{\frac{n}{d}}\cdots \sum _{l_k=1}^{\frac{n}{d}}(gcd(l_1,l_2,\dots ,l_k)=1) \\ =\sum _{d=1}^n{d}^2\sum _{l_1=1}^{\frac{n}{d}}\sum _{l_2=1}^{\frac{n}{d}}\cdots \sum _{l_k=1}^{\frac{n}{d}}\sum _{t\mid gcd(l_1,l_2,\dots ,l_k)}\mu (t) \\ =\sum _{d=1}^n{d}^2\sum _{t=1}^{\frac{n}{d}}\mu (t)(\frac{n}{td}{)}^2 \\ 另T=td \\ =\sum _{T=1}^n(\frac{n}{T}{)}^k\sum _{d\mid T}{d}^2\mu (\frac{T}{d}) \\ \sum _{i=2}^n{F}_n(i)=\sum _{T=1}^n\sum _{i=2}^k(\frac{n}{T}{)}^i\sum _{d\mid T}{d}^2\mu (\frac{T}{d}) \\ 到这一步前面的一部分只要对等比数列求和加上欧拉降幂就行 \\ 后面是一个积性函数前缀和，我们可以考虑通过杜教筛求解。 \\ f(n)=\sum _{d\mid n}{d}^2\mu (\frac{n}{d}) \\ f(n)=(\mu \ast i{d}^2)(n) \\ (f\ast I)(n)=(\mu \ast I\ast i{d}^2)(n)=i{d}^2(n) \\ \sum _{i=1}^n{i}^2=\sum _{i=1}^n\sum _{d\mid i}f(d)=\sum _{i=1}^n\sum _{d=1}^{\frac{n}{i}}f(d)=\sum _{i=1}^{n}S(\frac{n}{i}) \\ S(n)=\sum _{i=1}^n{i}^2-\sum _{i=2}^{n}S(\frac{n}{i}) \\ 还是写一下等比数列的求和公式吧 \\ \sum _{T=1}^n\sum _{i=2}^k\frac{\frac{n}{T}((\frac{n}{T}{)}^k-1)}{\frac{n}{T}-1}-\frac{n}{T} \\ 然后注意特判一下公比为1的特殊情况，因为这个给wa了一发。 \end{gathered}$$

代码

/*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 = 1e6 + 10, mod = 1e9 + 7, inv6 = 166666668;ll s[N];int prime[N], mu[N], cnt;bool st[N];ll quick_pow(ll a, ll n) {ll ans = 1;while(n) {if(n & 1) ans = ans * a % mod;a = a * a % mod;n >>= 1;}return ans;
}void init() {mu[1] = 1;for(int i = 2; i < N; i++) {if(!st[i]) {mu[i] = -1;prime[cnt++] = i;}for(int j = 0; j < cnt && i * prime[j] < N; j++) {st[i * prime[j]] = 1;if(i % prime[j] == 0) break;mu[i * prime[j]] = -mu[i];}}// for(int i = 1; i <= 10; i++) {//     cout << mu[i] << " \n"[i == 10];// }for(int i = 1; i < N; i++) {for(int j = i; j < N; j += i) {s[j] = (s[j] + 1ll * i * i % mod * mu[j / i] % mod + mod) % mod;}}// for(int i = 1; i <= 10; i++) {//     cout << s[i] << " \n"[i == 10];// }for(int i = 1; i < N; i++) {s[i] = (s[i] + s[i - 1]) % mod;}
}unordered_map<int, int> ans_s;ll S(ll n) {if(n < N) return s[n];if(ans_s.count(n)) return ans_s[n];ll ans = 1ll * n * (n + 1) % mod * (2 * n + 1) % mod * inv6 % mod;for(ll l = 2, r; l <= n; l = r + 1) {r = n / (n / l);ans = (ans - (r - l + 1) * S(n / l) % mod + mod) % mod;}return ans_s[n] = ans;
}ll calc(ll q, ll n, ll x) {if(q == 1) return (x - 1 + mod) % mod;ll ans = q * (quick_pow(q, n) - 1) % mod * quick_pow(q - 1, mod - 2) % mod;return (ans - q + mod) % 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; cin >> T;while(T--) {int n, k1 = 0, k2 = 0, sz; cin >> n;string str; cin >> str; sz = str.size();for(int i = 0; i < sz; i++) {k1 = (1ll * k1 * 10 + (str[i] - '0')) % (mod - 1);k2 = (1ll * k2 * 10 + (str[i] - '0')) % mod;}//公比为1的时候特判，所以记录两个k。// cout << n << " " << str << endl;ll ans = 0;for(ll l = 1, r; l <= n; l = r + 1) {r = n / (n / l);// cout << l << " " << r << endl;// cout << S(r) << " " << S(l - 1) << endl;ans = (ans + calc(n / l, k1, k2) * (S(r) - S(l - 1)) % mod + mod) % mod;// cout << ans << endl;}cout << ans << endl;// cout << endl;}return 0;
}