提交记录 28684


用户 题目 状态 得分 用时 内存 语言 代码长度
Gold_Dino 1002i. 【模板题】多项式乘法 Accepted 100 68.015 ms 7712 KB C++14 2.26 KB
提交时间 评测时间
2025-11-18 09:12:06 2025-11-18 09:12:11

#include <algorithm>
#include <vector>
#include <cstdio>
#include <string.h>
#include <random>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const ll Q = 998244353, G = 3;
const int MaxN = 6e5 + 7;
ll qpow(ll a,int n)
{
    ll bas = 1;
    while (n)
    {
        if (n & 1)
            (bas *= a) %= Q;
        (a *= a) %= Q, n >>= 1;
    }
    return bas;
}
ll inv(ll a) { return qpow(a, Q - 2); }
const ll ivG = inv(G);
int tr[MaxN];
inline void tpre(int len)
{
    for(int i = 1; i < len; ++i)
    {
        tr[i] = tr[i >> 1] >> 1;
        if (i & 1)
            tr[i] |= len >> 1;
    }
}
template<class Tp>
inline void clr(Tp *f, int len) { memset(f, 0, sizeof(Tp) * len); }
template<class Tp>
inline void cpy(Tp *f, Tp *g, int len) { memcpy(f, g, sizeof(Tp) * len); }
void NTT(int *f, bool tp, int len)
{
    static ull t1[MaxN], t2[MaxN];
    t2[0] = 1;
    tpre(len);
    for (int i = 0; i < len; ++i)
        t1[i] = f[tr[i]];
    for (int i = 1; i < len; i <<= 1)
    {
        ull wn = qpow(tp ? G : ivG, (Q - 1) / (i + i));
        for (int j = 1; j < i; ++j)
            t2[j] = t2[j - 1] * wn % Q;
        for (int j = 0; j < len; j += i)
            for (int k = 0; k < i; ++k)
            {
                ull t = t1[i | j | k] * t2[k] % Q;
                t1[i | j | k] = t1[j | k] + Q - t;
                t1[j | k] += t;
            }
        if (i == (1 << 10))
            for (int j = 0; j < len; ++j)
                t1[j] %= Q;
    }
    if (tp)
        for (int i = 0; i < len; ++i)
            f[i] = t1[i] % Q;
    else
    {
        ull ivl = inv(len);
        for (int i = 0; i < len; ++i)
            f[i] = t1[i] % Q * ivl % Q;
    }
}
void vmul(int *f, int *g, int len)
{
    for (int i = 0; i < len; ++i)
        f[i] = 1ll * f[i] * g[i] % Q;
}
void times(int *f, int *g, int m1, int m2)
{
    int len = 1 << (32 - __builtin_clz(m1 + m2 - 2));
    clr(f + m1, len - m1), clr(g + m2, len - m2);
    NTT(f, 1, len), NTT(g, 1, len);
    vmul(f, g, len);
    NTT(f, 0, len);
}
int f[MaxN], g[MaxN];
int main()
{
    int n, m, i;
    scanf("%d%d", &n, &m), ++n, ++m;
    for (i = 0; i < n; ++i) scanf("%d", f+i);
    for (i = 0; i < m; ++i) scanf("%d", g+i);
    times(f, g, n, m);
    for (i = 0; i < n+m-1; ++i)
        printf("%d ", f[i]);
    return 0;
}

CompilationN/AN/ACompile OKScore: N/A

Subtask #1 Testcase #115.63 us36 KBAcceptedScore: 100

Subtask #1 Testcase #267.478 ms7 MB + 464 KBAcceptedScore: 0

Subtask #1 Testcase #330.509 ms3 MB + 316 KBAcceptedScore: 0

Subtask #1 Testcase #430.482 ms3 MB + 304 KBAcceptedScore: 0

Subtask #1 Testcase #512.47 us36 KBAcceptedScore: 0

Subtask #1 Testcase #611.74 us36 KBAcceptedScore: 0

Subtask #1 Testcase #711.26 us36 KBAcceptedScore: 0

Subtask #1 Testcase #861.701 ms7 MB + 196 KBAcceptedScore: 0

Subtask #1 Testcase #961.757 ms7 MB + 196 KBAcceptedScore: 0

Subtask #1 Testcase #1056.054 ms6 MB + 952 KBAcceptedScore: 0

Subtask #1 Testcase #1167.734 ms7 MB + 544 KBAcceptedScore: 0

Subtask #1 Testcase #1268.015 ms6 MB + 424 KBAcceptedScore: 0

Subtask #1 Testcase #1311.46 us36 KBAcceptedScore: 0


Judge Duck Online | 评测鸭在线
Server Time: 2025-11-18 16:19:14 | Loaded in 1 ms | Server Status
个人娱乐项目,仅供学习交流使用 | 捐赠