题意：

求\(\sum_{i=1}^n\varphi(i)\)和\(\sum_{i=1}^n\mu(i)\)

思路：

由性质可知：\(\mu*I=\epsilon,\varphi*I=id\)那么可得：
\[ S_{\varphi}(n)=\sum_{i=1}^n\varphi(i)=\frac{(n+1)n}{2}-\sum_{i=2}^nS_{\varphi}(\lfloor\frac{n}{i}\rfloor)\\ S_{\mu}(n)\sum_{i=1}^n\mu(i)=1-\sum_{i=2}^nS_{\mu}(\lfloor\frac{n}{i}\rfloor) \]
然后用现预处理一部分答案，然后数论分块其他答案，用记忆化记录中间答案。
很卡常...

还有一种不用\(map\)的方法，详见代码\(2\)。

代码：

#include<map>
#include<set>
#include<cmath>
#include<cstdio>
#include<stack>
#include<ctime>
#include<vector>
#include<queue>
#include<cstring>
#include<string>
#include<sstream>
#include<iostream>
#include<algorithm>
#include<unordered_map>
typedef long long ll;
typedef unsigned long long ull;
using namespace std;
const int maxn = 6000000 + 5;
const ll MOD = 998244353;
const ull seed = 131;
const int INF = 0x3f3f3f3f;int vis[maxn];
int prime[maxn], cnt;
int N = 6000000;
ll mu[maxn], phi[maxn];
unordered_map<int, int> mmu;
unordered_map<int, ll> mphi;
void init(int n){cnt = 0;mu[1] = 1;phi[1] = 1;for(int i = 2; i <= n; i++){if(!vis[i]){prime[cnt++] = i;mu[i] = -1;phi[i] = i - 1;}for(int j = 0; j < cnt && prime[j] * i <= n; j++){vis[i * prime[j]] = 1;if(i % prime[j] == 0){mu[i * prime[j]] = 0;phi[i * prime[j]] = phi[i] * prime[j];break;}mu[i * prime[j]] = -mu[i];phi[i * prime[j]] = phi[i] * (prime[j] - 1);}}for(int i = 2; i <= n; i++){phi[i] += phi[i - 1];mu[i] += mu[i - 1];}
}ll Sphi(int n){if(n <= N) return phi[n];if(mphi.count(n)) return mphi[n];ll ans = 1LL * n * (1 + n) >> 1LL;for(int l = 2, r; l <= n; l = r + 1){r = min(n / (n / l), n);ans -= (r - l + 1) * Sphi(n / l);}return mphi[n] = ans;
}
int Smu(int n){if(n <= N) return mu[n];if(mmu.count(n)) return mmu[n];int ans = 1;for(int l = 2, r; l <= n; l = r + 1){r = min(n / (n / l), n);ans -= (r - l + 1) * Smu(n / l);}return mmu[n] = ans;
}int main(){init(N);int T;scanf("%d", &T);while(T--){int n;scanf("%d", &n);printf("%lld %d\n", Sphi(n), Smu(n));}return 0;
}

#include<map>
#include<set>
#include<cmath>
#include<cstdio>
#include<stack>
#include<ctime>
#include<vector>
#include<queue>
#include<cstring>
#include<string>
#include<sstream>
#include<iostream>
#include<algorithm>
typedef long long ll;
typedef unsigned long long ull;
using namespace std;
const int maxn = 6000000 + 5;
const ll MOD = 998244353;
const ull seed = 131;
const int INF = 0x3f3f3f3f;
int vis[maxn];
int prime[maxn], cnt;
int N = 6000000;
ll phi[maxn], mphi[100000];
int mu[maxn], mmu[100000];
int vis2[100000];
int n, up;
void init(int n){cnt = 0;mu[1] = 1;phi[1] = 1;for(int i = 2; i <= n; i++){if(!vis[i]){prime[cnt++] = i;mu[i] = -1;phi[i] = i - 1;}for(int j = 0; j < cnt && prime[j] * i <= n; j++){vis[i * prime[j]] = 1;if(i % prime[j] == 0){mu[i * prime[j]] = 0;phi[i * prime[j]] = phi[i] * prime[j];break;}mu[i * prime[j]] = -mu[i];phi[i * prime[j]] = phi[i] * (prime[j] - 1);}}for(int i = 2; i <= n; i++){phi[i] += phi[i - 1];mu[i] += mu[i - 1];}
}
int getmu(int x){return x <= N? mu[x] : mmu[up / x];
}
ll getphi(int x){return x <= N? phi[x] : mphi[up / x];
}
void solve(int n){int t = up / n;if(n <= N) return;if(vis2[t]) return;vis2[t] = 1;mmu[t] = 1, mphi[t] = 1LL * n * (n + 1) / 2;for(int l = 2, r; l <= n; l = r + 1){r = n / (n / l);solve(n / l);mmu[t] -= (r - l + 1) * getmu(n / l);mphi[t] -= (r - l + 1) * getphi(n / l);}
}int main(){init(N);int T;scanf("%d", &T);while(T--){scanf("%d", &n);up = n;if(n <= N) printf("%lld %d\n", phi[n], mu[n]);else{memset(vis2, 0, sizeof(vis2));solve(n);printf("%lld %d\n", mphi[1], mmu[1]);}}return 0;
}