1 条题解
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0

#include <bits/stdc++.h> #define lg2 std::__lg using std::cin; using std::cout; typedef unsigned long long u64; const int N = 1050000, mod = 998244353, iv2 = (mod + 1) / 2, unity = 31; typedef int vec[N], *pvec; vec inv, fact, finv; inline int min(const int x, const int y) {return x < y ? x : y;} inline int max(const int x, const int y) {return x < y ? y : x;} inline int & reduce(int &x) {return x += x >> 31 & mod;} inline int & neg(int &x) {return x = (!x - 1) & (mod - x);} u64 PowerMod(u64 a, int n, u64 c = 1) {for (; n; n >>= 1, a = a * a % mod) if (n & 1) c = c * a % mod; return c;} namespace poly_base { int l, n; u64 iv; vec w2; void init(int n = N, bool dont_calc_factorials = true) { int i, t; for (inv[1] = 1, i = 2; i < n; ++i) inv[i] = u64(mod - mod / i) * inv[mod % i] % mod; if (!dont_calc_factorials) for (*finv = *fact = i = 1; i < n; ++i) fact[i] = (u64)fact[i - 1] * i % mod, finv[i] = (u64)finv[i - 1] * inv[i] % mod; t = min(n > 1 ? lg2(n - 1) : 0, 21), *w2 = 1, w2[1 << t] = PowerMod(unity, 1 << (21 - t)); for (i = t; i; --i) w2[1 << (i - 1)] = (u64)w2[1 << i] * w2[1 << i] % mod; for (i = 1; i < n; ++i) w2[i] = (u64)w2[i & (i - 1)] * w2[i & -i] % mod; } inline void NTT_init(int len) {n = 1 << (l = len), iv = mod - (mod - 1) / n;} void DIF(int *a) { int i, *j, *k, len = n >> 1, R, *o; for (i = 0; i < l; ++i, len >>= 1) for (j = a, o = w2; j != a + n; j += len << 1, ++o) for (k = j; k != j + len; ++k) R = (u64)*o * k[len] % mod, reduce(k[len] = *k - R), reduce(*k += R - mod); } void DIT(int *a) { int i, *j, *k, len = 1, R, *o; for (i = 0; i < l; ++i, len <<= 1) for (j = a, o = w2; j != a + n; j += len << 1, ++o) for (k = j; k != j + len; ++k) reduce(R = *k + k[len] - mod), k[len] = u64(*k - k[len] + mod) * *o % mod, *k = R; } inline void DNTT(int *a) {DIF(a);} inline void IDNTT(int *a) { DIT(a), std::reverse(a + 1, a + n); for (int i = 0; i < n; ++i) a[i] = a[i] * iv % mod; } inline void DIF(int *a, int *b) {memcpy(b, a, n << 2), DIF(b);} inline void DIT(int *a, int *b) {memcpy(b, a, n << 2), DIT(b);} inline void DNTT(int *a, int *b) {memcpy(b, a, n << 2), DNTT(b);} inline void IDNTT(int *a, int *b) {memcpy(b, a, n << 2), IDNTT(b);} } int pn = 0, c[500000], p[41554], d[N], de[500000]; vec f; void sieve(int n) { int i, j, v; d[1] = 1; memset(c, -1, sizeof c); for (i = 2; i <= n; ++i) { if (!~c[i]) p[pn] = i, c[i] = pn++, d[i] = 2, de[i] = 1; for (j = 0; (v = i * p[j]) <= n && j < c[i]; ++j) c[v] = j, d[v] = (de[v] = d[i]) * 2; if (v <= n) c[v] = j, d[v] = d[i] + de[i], de[v] = de[i]; } } void work() { int i; namespace pb = poly_base; sieve(499999), pb::init(), pb::NTT_init(20), pb::DIF(d); for (i = 0; i < pb::n; ++i) d[i] = (u64)d[i] * d[i] % mod; pb::DIT(d); for (i = 2; i <= 500000; ++i) f[i] = d[pb::n - i] * pb::iv % mod; } int main() { int i, l, r, q; std::ios::sync_with_stdio(false), cin.tie(NULL); work(); for (cin >> q; q; --q) cin >> l >> r, i = std::max_element(f + l, f + (r + 1)) - f, cout << i << ' ' << f[i] << '\n'; return 0; }
- 1
信息
- ID
- 5780
- 时间
- 5000ms
- 内存
- 1028MiB
- 难度
- 10
- 标签
- 递交数
- 1
- 已通过
- 1
- 上传者