Battlestation Operational

$$\begin{gathered} \sum _{i=1}^n\sum _{j=1}^i?\frac{i}{j}?[gcd(i,j)=1] \\ \sum _{i=1}^n\sum _{j=1}^i(\frac{i}{j}+1)[gcd(i,j)=1]?\sum _{i=1}^n\sum _{j=1}^i[gcd(i,j)=1,j\mid i] \\ \sum _{i=1}^n\sum _{j=1}^i(\frac{i}{j}+1)[gcd(i,j)=1]?n \\ \sum _{i=1}^n\sum _{j=1}^i\frac{i}{j}[gcd(i,j)=1]+\sum _{i=1}^n\sum _{j=1}^i[gcd(i,j)=1]?n \\ \sum _{i=1}^n\sum _{j=1}^i\frac{i}{j}[gcd(i,j)=1]+\sum _{i=1}?(i)?n \\ \sum _{i=1}^n\sum _{j=1}^i\frac{i}{j}[gcd(i,j)=1] \\ \sum _{d=1}^n\mu (d)\sum _{i=1}^{\frac{n}{d}}\sum _{j=1}^i\frac{i}{j} \\ 求解\sum _{i=1}^n\sum _{j=1}^i\frac{i}{j}=\sum _{i=1}^n\sum _{j=i}^n\frac{j}{i} \\ 可以發(fā)現(xiàn)這個東西具有區(qū)間分塊性質(zhì) \\ 所以可以利用差分然后求前綴和得到，具體實(shí)現(xiàn)看代碼 \\ 整體復(fù)雜度O(n\log n+T\times \sqrt{n}) \end{gathered}$$

/*Author : lifehappy */ #include <bits/stdc++.h>using namespace std;typedef long long ll;const int N = 1e6 + 10, mod = 1e9 + 7;int prime[N], phi[N], mu[N], f[N], cnt, n;bool st[N];void init() {mu[1] = phi[1] = 1;for(int i = 2; i < N; i++) {if(!st[i]) {prime[++cnt] = i;mu[i] = -1;phi[i] = i - 1;}for(int j = 1; j <= cnt && 1ll * i * prime[j] < N; j++) {st[i * prime[j]] = 1;if(i % prime[j] == 0) {phi[i * prime[j]] = phi[i] * prime[j];break;}phi[i * prime[j]] = phi[i] * (prime[j] - 1);mu[i * prime[j]] = -mu[i];}}for(int i = 1; i < N; i++) {for(int j = i; j < N; j += i) {int l = j, r = j + i;f[l] = (f[l] + (j / i)) % mod;if(r < N) f[r] = (f[r] - (j / i) + mod) % mod;}}for(int i = 1; i < N; i++) {phi[i] = (phi[i] + phi[i - 1]) % mod;mu[i] = (mu[i] + mu[i - 1] + mod) % mod;f[i] = (f[i] + f[i - 1]) % mod;}for(int i = 1; i < N; i++) {f[i] = (f[i] + f[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();while(scanf("%d", &n) != EOF) {ll ans = 0;for(int l = 1, r; l <= n; l = r + 1) {r = n / (n / l);ans = (ans + 1ll * (mu[r] - mu[l - 1]) * f[n / l] % mod + mod) % mod;}ans = (ans + phi[n] - n + mod) % mod;printf("%lld\n", ans);}return 0; }

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