1 条题解
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#include <bits/stdc++.h> #define N 133333 #define K 254 #define lg2(x) (31 - __builtin_clz(x)) typedef int vec[N], *pvec; typedef long long ll; const ll mod = 998244353, half_mod = (mod + 1) / 2, root = 31; const char output[30] = "0123456789ABCDEFGHIJKLMNOPQRS"; vec fact, finv, ln; ll PowerMod(ll a, int n, ll c = 1) {for (; n; n >>= 1, a = a * a % mod) if (n & 1) c = c * a % mod; return c;} namespace Poly { int l, n; vec rev, x, y; void NTT_init(int len) { if (l == len) return; n = 1 << (l = len); ll g = PowerMod(root, 1 << (23 - l)); *x = 1, *rev = 0; for (int i = 1; i < n; ++i) x[i] = x[i - 1] * g % mod, rev[i] = rev[i >> 1] >> 1 | (i & 1) << (l - 1); } void DNTT(int *d, int *t) { int i, *j, *k, len = 1, delta = n, R; for (i = 0; i < n; ++i) t[rev[i]] = d[i]; for (i = 0; i < l; ++i) { delta >>= 1; for (k = x, j = y; j < y + len; k += delta, ++j) *j = *k; for (j = t; j < t + n; j += len << 1) for (k = j; k < j + len; ++k) { R = (ll)y[k - j] * k[len] % mod; k[len] = (*k - R < 0 ? *k - R + mod : *k - R); *k = (*k + R >= mod ? *k + R - mod : *k + R); } len <<= 1; } } vec B1; void Mul(int deg, pvec a, pvec b, pvec c) { if (!deg) {*c = (ll)*a * *b % mod; return;} NTT_init(lg2(deg) + 1); int i; ll iv = PowerMod(n, mod - 2); DNTT(a, c); DNTT(b, B1); for (i = 0; i < n; ++i) B1[i] = (ll)B1[i] * c[i] % mod; DNTT(B1, c); std::reverse(c + 1, c + n); for (i = 0; i < n; ++i) c[i] = c[i] * iv % mod; } } char n[K]; int p, g, ALL = 0; int len = 0, dn[K]; vec f, fc, fr, ans; void init() { int i; ll c; for (*fact = i = 1; i < p; ++i) fact[i] = (ll)fact[i - 1] * i % p; for (--i, finv[i] = i; i; --i) finv[i - 1] = (ll)finv[i] * i % p; for (g = 0; i != p - 1; ) for (i = 1, c = ++g; c != 1; ++i, c = c * g % p); for (i = 0, c = 1; i != p - 1; ++i, c = c * g % p) ln[c] = i; } int C(int n, int r) {return n < r ? 0 : (ll)fact[n] * finv[r] % p * finv[n - r] % p;} void solve() { int i, j, ind; memset(f, 0, p << 2), *f = 1; for (j = 0; j < len; ++j) { memset(fc, 0, p << 3); for (i = 0; i <= dn[j]; ++i) ++fc[ ind = ln[C(dn[j], i)] ]; Poly::Mul(2 * (p - 1), f, fc, fr); for (i = 0; i < p - 1; ++i) f[i] = (fr[i] + fr[i + p - 1]) % 29; } for (i = 1; i < p; ++i) ALL -= ans[i] = f[ln[i]]; ++ALL %= 29, *ans = ALL + (ALL >> 31 & 29); } void decomposition(const char *s, int *ret) { int i, n = strlen(s), a[K]; for (i = 0; i < n; ++i) a[n - i - 1] = s[i] & 15, ALL = (ALL * 10 + a[n - i - 1]) % 29; for (; ; ) { for (i = n - 1; i; --i) a[i - 1] += a[i] % p * 10, a[i] /= p; for (; n && !a[n - 1]; --n); if (n) ret[len++] = a[0] % p, a[0] /= p; else return; } } int main() { scanf("%s%d", n, &p), init(), decomposition(n, dn), solve(); for (int i = 0; i < p; ++i) putchar(output[ans[i]]); return putchar(10), 0; }
- 1
信息
- ID
- 4294
- 时间
- 1000ms
- 内存
- 128MiB
- 难度
- 10
- 标签
- 递交数
- 1
- 已通过
- 1
- 上传者