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#include "nt/primesum_mod4.hpp"
#include "nt/primesum.hpp" #include "nt/primetable.hpp" template <typename T> struct PrimeSum_Mod_4 { ll N; ll sqN; PrimeSum<T> A, B; PrimeSum_Mod_4(ll N) : N(N), sqN(sqrtl(N)), A(N), B(N) {} pair<T, T> operator[](ll x) { T a = A[x], b = B[x]; return {(a + b) / T(2), (a - b) / T(2)}; } void calc_count() { A.calc([](ll x) -> T { return (x + 1) / 2; }); B.calc([](ll x) -> T { return ((x + 3) % 4 <= 1 ? 1 : 0); }); } void calc_sum() { A.calc([](ll x) -> T { ll n = (x + 1) / 2; return T(n) * T(n); }); B.calc([](ll x) -> T { ll n = (x + 1) / 2; return (n % 2 == 0 ? T(-n) : T(n)); }); } };
#line 2 "nt/primetable.hpp" template <typename T = int> vc<T> primetable(int LIM) { ++LIM; const int S = 32768; static int done = 2; static vc<T> primes = {2}, sieve(S + 1); if (done < LIM) { done = LIM; primes = {2}, sieve.assign(S + 1, 0); const int R = LIM / 2; primes.reserve(int(LIM / log(LIM) * 1.1)); vc<pair<int, int>> cp; for (int i = 3; i <= S; i += 2) { if (!sieve[i]) { cp.eb(i, i * i / 2); for (int j = i * i; j <= S; j += 2 * i) sieve[j] = 1; } } for (int L = 1; L <= R; L += S) { array<bool, S> block{}; for (auto& [p, idx]: cp) for (int i = idx; i < S + L; idx = (i += p)) block[i - L] = 1; FOR(i, min(S, R - L)) if (!block[i]) primes.eb((L + i) * 2 + 1); } } int k = LB(primes, LIM + 1); return {primes.begin(), primes.begin() + k}; } #line 2 "nt/primesum.hpp" /* N と完全乗法的関数 f の prefix sum 関数 F を与える。 n = floor(N/d) となる n に対する sum_{p <= n} f(p) を計算する。 特に、素数の k 乗和や、mod m ごとでの素数の k 乗和が計算できる。 Complexity: O(N^{3/4}/logN) time, O(N^{1/2}) space. */ template <typename T> struct PrimeSum { ll N; ll sqN; vc<T> sum_lo, sum_hi; bool calculated; PrimeSum(ll N) : N(N), sqN(sqrtl(N)), calculated(0) {} // [1, x] ただし、x = floor(N, i) の形 T operator[](ll x) { assert(calculated); return (x <= sqN ? sum_lo[x] : sum_hi[double(N) / x]); } template <typename F> void calc(const F f) { auto primes = primetable<int>(sqN); sum_lo.resize(sqN + 1); sum_hi.resize(sqN + 1); FOR3(i, 1, sqN + 1) sum_lo[i] = f(i) - 1; FOR3(i, 1, sqN + 1) sum_hi[i] = f(double(N) / i) - 1; for (int p: primes) { ll pp = ll(p) * p; if (pp > N) break; int R = min(sqN, N / pp); int M = sqN / p; T x = sum_lo[p - 1]; T fp = sum_lo[p] - sum_lo[p - 1]; for (int i = 1; i <= M; ++i) sum_hi[i] -= fp * (sum_hi[i * p] - x); for (int i = M + 1; i <= R; ++i) sum_hi[i] -= fp * (sum_lo[N / (double(i) * p)] - x); for (int n = sqN; n >= pp; --n) sum_lo[n] -= fp * (sum_lo[n / p] - x); } calculated = 1; } void calc_count() { calc([](ll x) -> T { return x; }); } void calc_sum() { calc([](ll x) -> T { ll a = x, b = x + 1; if (!(x & 1)) a /= 2; if (x & 1) b /= 2; return T(a) * T(b); }); } }; #line 3 "nt/primesum_mod4.hpp" template <typename T> struct PrimeSum_Mod_4 { ll N; ll sqN; PrimeSum<T> A, B; PrimeSum_Mod_4(ll N) : N(N), sqN(sqrtl(N)), A(N), B(N) {} pair<T, T> operator[](ll x) { T a = A[x], b = B[x]; return {(a + b) / T(2), (a - b) / T(2)}; } void calc_count() { A.calc([](ll x) -> T { return (x + 1) / 2; }); B.calc([](ll x) -> T { return ((x + 3) % 4 <= 1 ? 1 : 0); }); } void calc_sum() { A.calc([](ll x) -> T { ll n = (x + 1) / 2; return T(n) * T(n); }); B.calc([](ll x) -> T { ll n = (x + 1) / 2; return (n % 2 == 0 ? T(-n) : T(n)); }); } };