結果
問題 | No.8120 Aoki's Present for Takahashi |
ユーザー |
![]() |
提出日時 | 2025-03-24 22:45:39 |
言語 | C++23 (gcc 13.3.0 + boost 1.87.0) |
結果 |
WA
(最新)
AC
(最初)
|
実行時間 | - |
コード長 | 4,194 bytes |
コンパイル時間 | 7,970 ms |
コンパイル使用メモリ | 333,612 KB |
実行使用メモリ | 29,812 KB |
最終ジャッジ日時 | 2025-04-01 23:11:39 |
合計ジャッジ時間 | 10,010 ms |
ジャッジサーバーID (参考情報) |
judge3 / judge1 |
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ファイルパターン | 結果 |
---|---|
sample | AC * 1 |
other | WA * 20 |
ソースコード
#include<bits/stdc++.h> #include<atcoder/all> using namespace std; using namespace atcoder; struct Barrett { using u32 = unsigned int; using i64 = long long; using u64 = unsigned long long; u32 m; u64 im; Barrett() : m(), im() {} Barrett(int n) : m(n), im(u64(-1) / m + 1) {} constexpr inline i64 quo(u64 n) { u64 x = u64((__uint128_t(n) * im) >> 64); u32 r = n - x * m; return m <= r ? x - 1 : x; } constexpr inline i64 rem(u64 n) { u64 x = u64((__uint128_t(n) * im) >> 64); u32 r = n - x * m; return m <= r ? r + m : r; } constexpr inline pair<i64, int> quorem(u64 n) { u64 x = u64((__uint128_t(n) * im) >> 64); u32 r = n - x * m; if (m <= r) return {x - 1, r + m}; return {x, r}; } constexpr inline i64 pow(u64 n, i64 p) { u32 a = rem(n), r = m == 1 ? 0 : 1; while (p) { if (p & 1) r = rem(u64(r) * a); a = rem(u64(a) * a); p >>= 1; } return r; } }; #define PRIME_POWER_BINOMIAL_M_MAX ((1LL << 30) - 1) #define PRIME_POWER_BINOMIAL_N_MAX 20000000 struct prime_power_binomial { int p, q, M; vector<int> fac, ifac, inv; int delta; Barrett bm, bp; prime_power_binomial(int _p, int _q) : p(_p), q(_q) { assert(1 < p && p <= PRIME_POWER_BINOMIAL_M_MAX); assert(_q > 0); long long m = 1; while (_q--) { m *= p; assert(m <= PRIME_POWER_BINOMIAL_M_MAX); } M = m; bm = Barrett(M), bp = Barrett(p); enumerate(); delta = (p == 2 && q >= 3) ? 1 : M - 1; } void enumerate() { int MX = min<int>(M, PRIME_POWER_BINOMIAL_N_MAX + 10); fac.resize(MX); ifac.resize(MX); inv.resize(MX); fac[0] = ifac[0] = inv[0] = 1; fac[1] = ifac[1] = inv[1] = 1; for (int i = 2; i < MX; i++) { if (i % p == 0) { fac[i] = fac[i - 1]; fac[i + 1] = bm.rem(1LL * fac[i - 1] * (i + 1)); i++; } else { fac[i] = bm.rem(1LL * fac[i - 1] * i); } } ifac[MX - 1] = bm.pow(fac[MX - 1], M / p * (p - 1) - 1); for (int i = MX - 2; i > 1; --i) { if (i % p == 0) { ifac[i] = bm.rem(1LL * ifac[i + 1] * (i + 1)); ifac[i - 1] = ifac[i]; i--; } else { ifac[i] = bm.rem(1LL * ifac[i + 1] * (i + 1)); } } } long long Lucas(long long n, long long m) { int res = 1; while (n) { int n0, m0; tie(n, n0) = bp.quorem(n); tie(m, m0) = bp.quorem(m); if (n0 < m0) return 0; res = bm.rem(1LL * res * fac[n0]); int buf = bm.rem(1LL * ifac[n0 - m0] * ifac[m0]); res = bm.rem(1LL * res * buf); } return res; } long long C(long long n, long long m) { if (n < m || n < 0 || m < 0) return 0; if (q == 1) return Lucas(n, m); long long r = n - m; int e0 = 0, eq = 0, i = 0; int res = 1; while (n) { res = bm.rem(1LL * res * fac[bm.rem(n)]); res = bm.rem(1LL * res * ifac[bm.rem(m)]); res = bm.rem(1LL * res * ifac[bm.rem(r)]); n = bp.quo(n); m = bp.quo(m); r = bp.quo(r); int eps = n - m - r; e0 += eps; if (e0 >= q) return 0; if (++i >= q) eq += eps; } if (eq & 1) res = bm.rem(1LL * res * delta); res = bm.rem(1LL * res * bm.pow(p, e0)); return res; } }; // constraints: // (M <= 1e7 and max(N) <= 1e18) or (M < 2^30 and max(N) <= 2e7) struct arbitrary_mod_binomial { long long mod; vector<int> M; vector<prime_power_binomial> cs; arbitrary_mod_binomial(long long md) : mod(md) { assert(1 <= md); assert(md <= PRIME_POWER_BINOMIAL_M_MAX); for (long long i = 2; i * i <= md; i++) { if (md % i == 0) { int j = 0, k = 1; while (md % i == 0) md /= i, j++, k *= i; M.push_back(k); cs.emplace_back(i, j); assert(M.back() == cs.back().M); } } if (md != 1) { M.push_back(md); cs.emplace_back(md, 1); } assert(M.size() == cs.size()); } long long C(long long n, long long m) { if (mod == 1) return 0; vector<long long> rem, d; for (int i = 0; i < (int)cs.size(); i++) { rem.push_back(cs[i].C(n, m)); d.push_back(M[i]); } return atcoder::crt(rem, d).first; } }; #define int long long signed main(){ cin.tie(0)->ios::sync_with_stdio(0); arbitrary_mod_binomial k(998243353); int TT,T; cin >> TT >> T; for(int ii=0;ii<T;ii++){ if(ii==TT-1){ cout << -1 << '\n'; continue; } int N,M; cin >> N >> M; //MCN cout << k.C(M,N) << '\n'; } }