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  1. #include <bits/stdc++.h>
  2. using namespace std;
  3.  
  4. #define ll long long
  5. #define all(c) ((c).begin()), ((c).end())
  6. #define sz(x) ((int)(x).size())
  7.  
  8. #ifdef LOCAL
  9. #include <print.h>
  10. #else
  11. #define trace(...)
  12. #define endl "\n" // remove in interactive
  13. #endif
  14. int power(int n, int p){
  15.     n /= p;
  16.     int ret = 0;
  17.     while(n){
  18.         ret += n;
  19.         n /= p;
  20.     }
  21.     return ret;
  22. }
  23.  
  24. int main(){
  25.     ios_base::sync_with_stdio(false); 
  26.     cin.tie(NULL); // Remove in interactive problems
  27.     int m; ll N;
  28.     cin >> N >> m;
  29.     m = min((ll)m, N);
  30.     vector<bool> isprime(m+1, true);
  31.  
  32.     for(int p = 2; p * p <= m; p++) if(isprime[p]){
  33.         for(int q = p * p; q <= m; q += p) isprime[q] = false;
  34.     }
  35.     vector<ll> Sum(m + 1);
  36.     vector<int> tot;
  37.     vector<ll> primes;
  38.     for(int p = 2; p <= m; p++){
  39.         Sum[p] = Sum[p - 1];
  40.         if(p % 3 == 2 && isprime[p]){
  41.             primes.push_back(p);
  42.             tot.push_back(power(m, p));
  43.             Sum[p] += p;
  44.         }
  45.     }
  46.     int nump = primes.size();
  47.     const int mod = 998244353;
  48.     const ll mod2 = mod * 8LL * mod;
  49.  
  50.     function<ll(ll)> get = [&](ll n){
  51.         ll ret = 2; // y = 1
  52.         function<void(ll, int, int,int)> recurse = [&](ll x, int i, int t, int e){
  53.             int p = x == 1 ? 1 : primes[i];
  54.             if(e > 0 && e < tot[i]) ret += t * x * p; // y = x * p
  55.             ret += ((3 - t) * x % mod) * ((Sum[min((ll)m, n / x)] - Sum[p]) % mod); // y = x * q, q > p
  56.             if(ret >= mod2) ret -= mod2;
  57.             ll mx = n / x;
  58.             if(primes[i] *  primes[i] > mx) return;
  59.             if(e < tot[i]) recurse(x * primes[i], i, e == 0 ? 3 - t: t, e + 1);
  60.             for(int j = i + 1; j < nump && primes[j] * primes[j] <= mx; j++){
  61.                 recurse(x * primes[j], j, 3 - t, 1);
  62.             }
  63.         };
  64.         recurse(1, 0, 2, 0);
  65.         return ret % mod;
  66.     };
  67.     ll ans = 0;
  68.     int maxp2 = power(m, 2);
  69.     for(ll x = 2, i = 1; x <= N && i <= maxp2; x *= 2, i++) ans += x / 2;
  70.     int maxp = power(m, 3);
  71.  
  72.     for(ll x = 3, i = 1; x <= N && i <= maxp; x *= 3, i++){
  73.         ans += (get(N / x)) * ( (x / 3) % mod);
  74.         ans %= mod;
  75.     }
  76.     cout << ans << endl;
  77. }
Success #stdin #stdout 0s 5300KB
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https://ideone.com/3FTNgF
language:
C++14 (gcc 8.3)
created:
36 days ago
visibility:
secret

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