結果
問題 | No.407 鴨等素数間隔列の数え上げ |
ユーザー |
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提出日時 | 2016-10-21 17:23:42 |
言語 | C++14 (gcc 13.3.0 + boost 1.87.0) |
結果 |
AC
|
実行時間 | 3 ms / 1,000 ms |
コード長 | 2,039 bytes |
コンパイル時間 | 1,487 ms |
コンパイル使用メモリ | 99,216 KB |
実行使用メモリ | 5,248 KB |
最終ジャッジ日時 | 2024-12-16 01:42:15 |
合計ジャッジ時間 | 2,793 ms |
ジャッジサーバーID (参考情報) |
judge5 / judge4 |
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ファイルパターン | 結果 |
---|---|
sample | AC * 5 |
other | AC * 31 |
ソースコード
#pragma GCC optimize ("O3")#pragma GCC target ("avx")#include <cstdio>#include <cassert>#include <cmath>#include <cstring>#include <iostream>#include <algorithm>#include <vector>#include <map>#include <set>#include <functional>#include <stack>#include <queue>#include <tuple>#define getchar getchar_unlocked#define putchar putchar_unlocked#define _rep(_1, _2, _3, _4, name, ...) name#define rep2(i, n) rep3(i, 0, n)#define rep3(i, a, b) rep4(i, a, b, 1)#define rep4(i, a, b, c) for (int i = int(a); i < int(b); i += int(c))#define rep(...) _rep(__VA_ARGS__, rep4, rep3, rep2, _)(__VA_ARGS__)using namespace std;using i8 = signed char;using i16 = signed short;using i64 = long long;using u8 = unsigned char;using u32 = unsigned;using u64 = unsigned long long;using f80 = long double;void solve() {const int L_MAX = 1e9;constexpr int sqrt_L = sqrt(L_MAX) + 10;static i64 pis[sqrt_L + 1], pil[sqrt_L + 1];static i64 psums[sqrt_L + 1], psuml[sqrt_L + 1];int N, L;while (~scanf("%d %d", &N, &L)) {// sublinearauto tri = [&](int n) { return i64(n) * (n + 1) / 2; };int M = L / (N - 1), v = sqrt(M);rep(i, 1, v + 1) pis[i] = i - 1, pil[i] = M / i - 1;rep(i, 1, v + 1) psums[i] = tri(i) - 1, psuml[i] = tri(M / i) - 1;rep(p, 2, v + 1) if (pis[p] > pis[p - 1]) {auto psum = psums[p - 1];auto pcnt = pis[p - 1];auto q = i64(p) * p;rep(i, 1, min(M / q, i64(v)) + 1) {i64 d = i64(i) * p;pil[i] -= (d <= v ? pil[d] : pis[M / d]) - pcnt;psuml[i] -= ((d <= v ? psuml[d] : psums[M / d]) - psum) * p;}for (int i = v; i >= q; --i) {pis[i] -= pis[i / p] - pcnt;psums[i] -= (psums[i / p] - psum) * p;}}if (M <= 1) psuml[1] = pil[1] = 0;i64 ans = pil[1] * (L + 1) - (N - 1) * psuml[1];printf("%lld\n", ans);}}int main() {clock_t beg = clock();solve();clock_t end = clock();fprintf(stderr, "%.3f sec\n", double(end - beg) / CLOCKS_PER_SEC);}