这个是一个处理积极性函数前缀和的东西，先把能处理的范围的函数值处理出来，然后数论分块，递归处理这些值，一点点缩小这个的范围，从而计算出函数值。
具体看代码就行了，不是很难。
洛谷的模板题是处理μ和φ的前缀和。
代码：
#include<iostream> #include<cstdio> #include<cmath> #include<ctime> #include<queue> #include<algorithm> #include<cstring> #include<unordered_map> using namespace std; #define duke(i,a,n) for(register int i = a;i <= n;i++) #define lv(i,a,n) for(register int i = a;i >= n;i--) #define clean(a) memset(a,0,sizeof(a)) const int INF = 1 << 30; typedef long long ll; typedef double db; template <class T> void read(T &x) { char c; bool op = 0; while(c = getchar(), c < '0' || c > '9') if(c == '-') op = 1; x = c - '0'; while(c = getchar(), c >= '0' && c <= '9') x = x * 10 + c - '0'; if(op) x = -x; } template <class T> void write(T x) { if(x < 0) putchar('-'), x = -x; if(x >= 10) write(x / 10); putchar('0' + x % 10); } const int N = 5e6 + 5; int miu[N + 2],flag[N + 2],pri[N + 5]; ll phi[N + 2],tot = 0; unordered_map <int,ll> ansmiu,ansphi; void init() { miu[1] = 1; phi[1] = 1; duke(i,2,N) { if(!flag[i]) { pri[++tot] = i; phi[i] = i - 1; miu[i] = -1; } for(int j = 1;j <= tot;j++) { if(i * pri[j] > N) break; flag[i * pri[j]] = 1; if(i % pri[j] == 0) { phi[i * pri[j]] = phi[i] * pri[j]; break; } miu[i * pri[j]] = -miu[i]; phi[i * pri[j]] = phi[i] * (pri[j] - 1); } } duke(i,2,N) { miu[i] += miu[i - 1]; phi[i] += phi[i - 1]; } } int t; ll phis(int n) { if(n <= N) { return phi[n]; } if(ansphi.count(n)) return ansphi[n]; ll ans = (1LL * n * (n + 1)) >> 1; for(int l = 2,r;l <= n;l = r + 1) { r = n / (n / l); ans -= (r - l + 1) * phis(n / l); } ansphi[n] = ans; return ans; } ll mius(int n) { if(n <= N) { return miu[n]; } if(ansmiu.count(n)) return ansmiu[n]; ll ans = 1; for(int l = 2,r;l <= n;l = r + 1) { r = n / (n / l); ans -= (r - l + 1) * mius(n / l); } ansmiu[n] = ans; return ans; } int main() { init(); read(t); while(t--) { int n; read(n); printf("%lld %lld\n",phis(n),mius(n)); } return 0; }