杜教筛

#include <algorithm>
#include <cstdio>
#include <cstring>
#include <map>
using namespace std;
const int maxn = 2000010;
long long T, n, pri[maxn], cur, mu[maxn], sum_mu[maxn];
bool vis[maxn];
map<long long, long long> mp_mu;

long long S_mu(long long x) {  // 求mu的前缀和
  if (x < maxn) return sum_mu[x];
  if (mp_mu[x]) return mp_mu[x];  // 如果map中已有该大小的mu值，则可直接返回
  long long ret = (long long)1;
  for (long long i = 2, j; i <= x; i = j + 1) {
    j = x / (x / i);
    ret -= S_mu(x / i) * (j - i + 1);
  }
  return mp_mu[x] = ret;  // 路径压缩，方便下次计算
}

long long S_phi(long long x) {  // 求phi的前缀和
  long long ret = (long long)0;
  long long j;
  for (long long i = 1; i <= x; i = j + 1) {
    j = x / (x / i);
    ret += (S_mu(j) - S_mu(i - 1)) * (x / i) * (x / i);
  }
  return (ret - 1) / 2 + 1;
}

int main() {
  scanf("%lld", &T);
  mu[1] = 1;
  for (int i = 2; i < maxn; i++) {  // 线性筛预处理mu数组
    if (!vis[i]) {
      pri[++cur] = i;
      mu[i] = -1;
    }
    for (int j = 1; j <= cur && i * pri[j] < maxn; j++) {
      vis[i * pri[j]] = true;
      if (i % pri[j])
        mu[i * pri[j]] = -mu[i];
      else {
        mu[i * pri[j]] = 0;
        break;
      }
    }
  }
  for (int i = 1; i < maxn; i++)
    sum_mu[i] = sum_mu[i - 1] + mu[i];  // 求mu数组前缀和
  while (T--) {
    scanf("%lld", &n);
    printf("%lld %lld\n", S_phi(n), S_mu(n));
  }
  return 0;
}

#include <cmath>
#include <cstdio>
#include <map>
using namespace std;
const int N = 5e6, NP = 5e6, SZ = N;
long long n, P, inv2, inv6, s[N];
int phi[N], p[NP], cnt, pn;
bool bp[N];
map<long long, long long> s_map;

long long ksm(long long a, long long m) {  // 求逆元用
  long long res = 1;
  while (m) {
    if (m & 1) res = res * a % P;
    a = a * a % P, m >>= 1;
  }
  return res;
}

void prime_work(int k) {  // 线性筛phi，s
  bp[0] = bp[1] = 1, phi[1] = 1;
  for (int i = 2; i <= k; i++) {
    if (!bp[i]) p[++cnt] = i, phi[i] = i - 1;
    for (int j = 1; j <= cnt && i * p[j] <= k; j++) {
      bp[i * p[j]] = 1;
      if (i % p[j] == 0) {
        phi[i * p[j]] = phi[i] * p[j];
        break;
      } else
        phi[i * p[j]] = phi[i] * phi[p[j]];
    }
  }
  for (int i = 1; i <= k; i++)
    s[i] = (1ll * i * i % P * phi[i] % P + s[i - 1]) % P;
}

long long s3(long long k) {  // 立方和
  return k %= P, (k * (k + 1) / 2) % P * ((k * (k + 1) / 2) % P) % P;
}

long long s2(long long k) {  // 平方和
  return k %= P, k * (k + 1) % P * (k * 2 + 1) % P * inv6 % P;
}

long long calc(long long k) {  // 计算S(k)
  if (k <= pn) return s[k];
  if (s_map[k]) return s_map[k];  // 对于超过pn的用map离散存储
  long long res = s3(k), pre = 1, cur;
  for (long long i = 2, j; i <= k; i = j + 1)
    j = k / (k / i), cur = s2(j),
    res = (res - calc(k / i) * (cur - pre) % P) % P, pre = cur;
  return s_map[k] = (res + P) % P;
}

long long solve() {
  long long res = 0, pre = 0, cur;
  for (long long i = 1, j; i <= n; i = j + 1) {
    j = n / (n / i);
    cur = calc(j);
    res = (res + (s3(n / i) * (cur - pre)) % P) % P;
    pre = cur;
  }
  return (res + P) % P;
}

int main() {
  scanf("%lld%lld", &P, &n);
  inv2 = ksm(2, P - 2), inv6 = ksm(6, P - 2);
  pn = (long long)pow(n, 0.666667);  // n^(2/3)
  prime_work(pn);
  printf("%lld", solve());
  return 0;
}  // 不要为了省什么内存把数组开小,会卡80