[LOJ572] Misaka Network 与求和 [杜教筛+Min25]

Posted 2022-07-08 qq62c30ac77b2a7

​​传送门​​

首先 g 可以整除分块，然后杜教筛处理 g 

考虑如何用 Min25 筛出 f，一个 d 有 f(d) ^ k 的贡献，当且仅当有一个 >= 它的质数在它前面

所有的方案数就是 >= 它的质数个数

于是 Min25 先处理出质数个数，然后加上合数贡献时算一下 f(d)

具体一点就是 

#include<bits/stdc++.h>
#define N 1000050
using namespace std;
typedef long long ll;
typedef unsigned int ui;
ll n, m, k, S, tot, prim[N], w[N], id1[N], id2[N]; bool isp[N];
ui f[N], phi[N], fk[N];
ll Id(ll x) return x <= S ? id1[x] : id2[n/x];  
ui power(ui a, ll b) ui ans = 1;
  for(;b;b>>=1) if(b&1) ans *= a; a *= a;
  return ans;
 
void prework()
  phi[1] = 1;
  for(int i = 2; i <= N-50; i++)
    if(!isp[i]) prim[++tot] = i, fk[tot] = power(i, k), phi[i] = i-1;
    for(int j = 1; j <= tot; j++)
      if(i * prim[j] > N-50) break;
      isp[i * prim[j]] = 1; if(i % prim[j] == 0) phi[i * prim[j]] = phi[i] * prim[j]; break;
      phi[i * prim[j]] = phi[i] * phi[prim[j]];
    
  
  for(int i = 2; i <= N-50; i++) phi[i] += phi[i-1];

ll Solve(ll x, int p)
  if(prim[p] > x) return 0; ll ans = 0;
  for(int i = p; i <= tot && prim[i] * prim[i] <= x; i++)
    for(ll e = 1, l = prim[i]; l * prim[i] <= x; e++, l *= prim[i])
      ans += Solve(x / l, i + 1) + fk[i] * (f[Id(x / l)] - (i-1));
    
   return ans;

ui Phi[N];
ui Getphi(ll x)
  if(x <= N-50) return phi[x];
  int k = Id(x);
  if(Phi[k]) return Phi[k];
  ui ans = 1ll * x * (x + 1) / 2;
  for(ll l = 2, r; l <= x; l = r+1)
    ll v = x/l; r = x/v;
    ans -= (r-l+1) * Getphi(v);
   return Phi[k] = ans;
 
int main()
  scanf("%lld%lld", &n, &k); m = 0; S = sqrt(n); prework();
  for(ll l = 1, r; l <= n; l = r+1)
    ll v = n/l; r = n/v; 
    if(v <= S) id1[v] = ++m; else id2[n/v] = ++m;
    w[m] = v; f[m] = w[m]-1;
  
  for(int i = 1; i <= tot; i++)
    if(prim[i] * prim[i] > n) break;
    for(int j = 1; j <= m && prim[i] * prim[i] <= w[j]; j++)
      f[j] -= f[Id(w[j]/prim[i])] - (i - 1);
    
   
  ui ans = 0;
  for(ll l = 1, r, lst = 0, nxt = 0; l <= n; l = r+1)
    ui v = n/l; r = n/v;
    ui v1 = 2 * Getphi(v) - 1;
    nxt = Solve(r, 1) + f[Id(r)];
    ans += (nxt - lst) * v1;
    lst = nxt;
   cout << ans; return 0;