function

推式子

$$\begin{gathered} n^2-3n+2=\sum _{d\mid n}f(d) \\ calulateS(n)\sum _{i=1}^{n}f(i)\mathrm{m}\mathrm{o}\mathrm{d}10^9+7 \end{gathered}$$

考虑对式子进行化简
$$\begin{gathered} \sum _{i=1}^n(i^2-3i+2)=\sum _{i=1}^n\sum _{d\mid i}f(d) \\ =\sum _{i=1}^n\sum _{d=1}^{\frac{n}{d}}f(d) \\ =\sum _{i=1}^{n}S(\frac{n}{d}) \\ =s(n)+\sum _{i=2}^{n}S(\frac{n}{d}) \end{gathered}$$
然后就可以进行杜教筛了，但是在杜教筛之前我们得先要预处理出前$10^6$个数的$S(i)$来，这里就可以通过$10^{6}lo{g}^2(10^6)$来得到。

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 mod = 1e9 + 7, N = 1e6 + 10, inv4 = 250000002, inv2 = 500000004, inv3 = (mod + 1) / 3;ll sum[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() {for(int i = 1; i < N; i++) sum[i] = 1ll * i * i - 3 * i + 2;for(int i = 1; i < N; i++) {for(int j = i + i; j < N; j += i) {sum[j] -= sum[i];}}// for(int i = 1; i <= 10; i++) {//     cout << sum[i] << " ";// }// puts("");for(int i = 1; i < N; i++) {sum[i] = (sum[i] + sum[i - 1]) % mod;}
}unordered_map<int, int> ans_s;ll calc(ll n) {n % mod;if(n < 2) n += mod;return 1ll * n * (n - 1) % mod * (n - 2) % mod * inv3 % mod;
}ll S(ll n) {if(n < N) return sum[n];if(ans_s.count(n)) return ans_s[n];ll ans = calc(n);for(ll l = 2, r; l <= n; l = r + 1) {r = n / (n / l);ans = (ans - (r - l + 1) % mod * S(n / l) % mod + mod) % mod;}ans_s[n] = ans;return ans;
}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--) {ll n = read();printf("%lld\n", S(n));}return 0;
}