提交记录 16893


用户 题目 状态 得分 用时 内存 语言 代码长度
Saisyc 1002i. 【模板题】多项式乘法 Accepted 100 38.291 ms 9632 KB C++ 2.51 KB
提交时间 评测时间
2021-10-26 21:20:57 2021-10-26 21:21:00
#include <cstdio>
#include <cctype>
#include <algorithm>

namespace fastIO {
    const int SZ = 1 << 25;
    char ibuf[SZ], *p0 = ibuf, *p1 = ibuf;
    char getchar(void) { return p0 == p1 && (p1 = (p0 = ibuf) + fread(ibuf, 1, SZ, stdin), p0 == p1) ? EOF : *p0++; }
    char char_read; void read(auto &val) {
		val = 0; do char_read = getchar(); while(!isdigit(char_read));
		do val = val * 10 + char_read - '0'; while(isdigit(char_read = getchar()));
    }
    char obuf[SZ], *p = obuf;
    void putchar(char char_write) { *p++ = char_write; }
    void write(auto val) { if(val >= 10) write(val / 10); putchar(val % 10 + '0'); }
    void write(auto a[], int sz) { for(int i = 0; i < sz; ++i) write(a[i]), putchar(' '); putchar('\n'); }
    void output(void) { fwrite(obuf, p - obuf, 1, stdout);}
};

const int N = 1 << 21;
const long long p = 998244353;
const long long q = 31596, qq = 62863;

inline long long mul_modp(long long a, long long b) {
//	long long x = a * b - (long long)((long double)a * b / p) * p; return x < p ? x : x - p;
	long long a0 = a % q, a1 = a / q;
	long long b0 = b % q, b1 = b / q;
	return (a0 * b0 + (a0 * b1 + a1 * b0) * q + a1 * b1 * qq) % p;
}

long long pow_modp(long long a, long long b) { long long x = 1; for(; b; a = mul_modp(a, a), b >>= 1) if(b & 1) x = mul_modp(x, a); return x; }
#define w(n) pow_modp(         3, (p - 1) / n)
#define v(n) pow_modp( 332748118, (p - 1) / n)

long long ntt_sup[N];
void ntt(auto a[], int n, long long w) {
	auto ntt_a = ntt_sup; for(int i = 1; i < n; i <<= 1, w = mul_modp(w, w), std :: swap(a, ntt_a)) {
		auto _w = 1; for(auto k0 = ntt_a, j0 = a, j1 = a + (n >> 1); k0 != ntt_a + n; k0 += i, _w = mul_modp(_w, w))
			for(auto k1 = k0 + i, _k0 = k1; k0 != _k0; ++j0, ++j1, ++k0, ++k1)
				*k0 = *j0 + *j1 < 0 ? *j0 + *j1 + p : *j0 + *j1 - p,
				*k1 = mul_modp(*j0 - *j1, _w);
	}
	if(ntt_a != ntt_sup) std :: copy(a, a + n, ntt_a);
}
void poly_mul(auto a[], auto b[], int n) {
	ntt(a, n, w(n)), ntt(b, n, w(n)); for(int i = 0; i < n; ++i) a[i] = mul_modp(a[i], b[i]); ntt(a, n, v(n));
	int inv_n = pow_modp(n, p - 2); for(int i = 0; i < n; ++i) a[i] = mul_modp(a[i], inv_n);
	for(int i = 0; i < n; ++i) a[i] = a[i] < 0 ? a[i] + p : a[i];
}

int n, n_a, n_b;
long long a[N], b[N];

int main(void) {
	fastIO :: read(n_a), fastIO :: read(n_b); ++n_a, ++n_b; for(n = 1; n < n_a + n_b - 1; n <<= 1);
	for(int i = 0; i < n_a; ++i) fastIO :: read(a[i]);
	for(int i = 0; i < n_b; ++i) fastIO :: read(b[i]);
	poly_mul(a, b, n); n = n_a + n_b - 1;
	fastIO :: write(a, n), fastIO :: output();
	return 0;
}

CompilationN/AN/ACompile OKScore: N/A

Subtask #1 Testcase #18.27 us32 KBAcceptedScore: 0

Subtask #1 Testcase #238.097 ms9 MB + 252 KBAcceptedScore: 100

Subtask #1 Testcase #317.259 ms3 MB + 788 KBAcceptedScore: 0

Subtask #1 Testcase #417.36 ms3 MB + 768 KBAcceptedScore: 0

Subtask #1 Testcase #58.95 us32 KBAcceptedScore: 0

Subtask #1 Testcase #69.28 us32 KBAcceptedScore: 0

Subtask #1 Testcase #78.23 us32 KBAcceptedScore: 0

Subtask #1 Testcase #837.294 ms8 MB + 676 KBAcceptedScore: 0

Subtask #1 Testcase #937.331 ms8 MB + 676 KBAcceptedScore: 0

Subtask #1 Testcase #1036.549 ms8 MB + 72 KBAcceptedScore: 0

Subtask #1 Testcase #1138.291 ms9 MB + 416 KBAcceptedScore: 0

Subtask #1 Testcase #1235.396 ms7 MB + 172 KBAcceptedScore: 0

Subtask #1 Testcase #138.26 us32 KBAcceptedScore: 0


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