[UOJ 34]多项式乘法

Description

这是一道模板题。

给你两个多项式，请输出乘起来后的多项式。

Input

第一行两个整数 $n$ 和 $m$，分别表示两个多项式的次数。

第二行 $n + 1$ 个整数，表示第一个多项式的 $0$ 到 $n$ 次项系数。

第三行 $m + 1$ 个整数，表示第二个多项式的 $0$ 到 $m$ 次项系数。

Output

一行 $n + m + 1$ 个整数，表示乘起来后的多项式的 $0$ 到 $n + m$ 次项系数。

Sample Input

1 2
1 2
1 2 1

Sample Output

1 4 5 2

Sample Explanation

$(1 + 2x) cdot (1 + 2x + x^2) = 1 + 4x + 5x^2 + 2x^3$。

HINT

$0 leq n, m leq 10^5$，保证输入中的系数大于等于 $0$ 且小于等于 $9$。

时间限制：$1 exttt{s}$

空间限制：$256 exttt{MB}$

题解

终于会背模板 $FFT$ 辣！！

//It is made by Awson on 2018.1.27
#include <set>
#include <map>
#include <cmath>
#include <ctime>
#include <queue>
#include <stack>
#include <cstdio>
#include <string>
#include <vector>
#include <cstdlib>
#include <cstring>
#include <complex>
#include <iostream>
#include <algorithm>
#define LL long long
#define dob complex<double>
#define Abs(a) ((a) < 0 ? (-(a)) : (a))
#define Max(a, b) ((a) > (b) ? (a) : (b))
#define Min(a, b) ((a) < (b) ? (a) : (b))
#define Swap(a, b) ((a) ^= (b), (b) ^= (a), (a) ^= (b))
#define writeln(x) (write(x), putchar('
'))
#define lowbit(x) ((x)&(-(x)))
using namespace std;
const int INF = ~0u>>1;
const double pi = acos(-1.0);
const int N = 1e5*4;
void read(int &x) {
    char ch; bool flag = 0;
    for (ch = getchar(); !isdigit(ch) && ((flag |= (ch == '-')) || 1); ch = getchar());
    for (x = 0; isdigit(ch); x = (x<<1)+(x<<3)+ch-48, ch = getchar());
    x *= 1-2*flag;
}
void write(int x) {
    if (x > 9) write(x/10);
    putchar(x%10+48);
}

int n, m, x, L, R[N+5];
dob a[N+5], b[N+5];

void FFT(dob *A, int o) {
    for (int i = 0; i < n; i++) if (i > R[i]) swap(A[i], A[R[i]]);
    for (int i = 1; i < n; i <<= 1) {
        dob wn(cos(pi/i), sin(pi*o/i)), x, y;
        for (int j = 0; j < n; j += (i<<1)) {
            dob w(1, 0);
            for (int k = 0; k < i; k++, w *= wn) {
                x = A[j+k], y = w*A[i+j+k];
                A[j+k] = x+y, A[i+j+k] = x-y;
            }
        }
    }
}
void work() {
    read(n), read(m);
    for (int i = 0; i <= n; i++) read(x), a[i] = x;
    for (int i = 0; i <= m; i++) read(x), b[i] = x;
    m += n;
    for (n = 1; n <= m; n <<= 1) L++;
    for (int i = 0; i < n; i++) R[i] = (R[i>>1]>>1)|((i&1)<<(L-1));
    FFT(a, 1), FFT(b, 1);
    for (int i = 0; i < n; i++) a[i] *= b[i];
    FFT(a, -1);
    for (int i = 0; i <= m; i++) write(int(a[i].real()/n+0.5)), putchar(' ');
} 
int main() {
    work();
    return 0;
}

upd 18.4.2：

$NTT$ 的板子：

#include <bits/stdc++.h>
using namespace std;
const int mod = 998244353;
const int N = 4e5;

int n, m, a[N+5], b[N+5], x, L, R[N+5];

int quick_pow(int a, int b) {
    int ans = 1;
    while (b) {
    if (b&1) ans = 1ll*ans*a%mod;
    b >>= 1, a = 1ll*a*a%mod;
    }
    return ans;
}
void NTT(int *A, int o) {
    for (int i = 0; i < n; i++) if (i < R[i]) swap(A[i], A[R[i]]);
    for (int i = 1; i < n; i <<= 1) {
    int gn = quick_pow(3, (mod-1)/(i<<1)), x, y;
    if (o == -1) gn = quick_pow(gn, mod-2);
    for (int j = 0; j < n; j += (i<<1)) {
        int g = 1;
        for (int k = 0; k < i; k++, g = 1ll*g*gn%mod) {
        x = A[j+k], y = 1ll*g*A[j+k+i]%mod;
        A[j+k] = (x+y)%mod;
        A[j+k+i] = (x-y+mod)%mod;
        }
    }
    }
}
void work() {
    scanf("%d%d", &n, &m);
    for (int i = 0; i <= n; i++) scanf("%d", &x), a[i] = x;
    for (int i = 0; i <= m; i++) scanf("%d", &x), b[i] = x;
    m += n;
    for (n = 1; n <= m; n <<= 1) ++L;
    for (int i = 0; i < n; i++) R[i] = (R[i>>1]>>1)|((i&1)<<(L-1));
    NTT(a, 1), NTT(b, 1);
    for (int i = 0; i < n; i++) a[i] = 1ll*a[i]*b[i]%mod;
    NTT(a, -1);
    int inv = quick_pow(n, mod-2);
    for (int i = 0; i <= m; i++) printf("%d ", int(1ll*a[i]*inv%mod));
}
int main() {work(); return 0; }