本文主要是介绍LG P3803 【模板】多项式乘法(FFT),对大家解决编程问题具有一定的参考价值,需要的程序猿们随着小编来一起学习吧!
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#include <cstdio>
#include <iostream>
#include <cmath>
#define re register
using namespace std;
const int N = 2e6 + 1e5;
int rev[N], n, m;
inline int read()
{
char ch = getchar(); int f = 1, x = 0;
while (ch < '0' || ch > '9') f = (ch == '-' ? -1 : f), ch = getchar();
while (ch >= '0' && ch <= '9') x = (x << 3) + (x << 1) + ch - '0', ch = getchar();
return x * f;
}
const double Pi = acos(-1.0);
struct complex{
double x, y;
inline complex operator + (const complex &a) const {return complex{x + a.x, y + a.y};}
inline complex operator - (const complex &a) const {return complex{x - a.x, y - a.y};}
inline complex operator * (const complex &a) const {return complex{x * a.x - y * a.y, x * a.y + y * a.x};}
}a[N], b[N];
inline void FFT(complex *a, int lim, int inv)
{
if (lim == 1) return;
for(re int i = 0; i < lim; i++)
if (i < rev[i]) swap(a[i], a[rev[i]]);
for(re int mid = 1; mid < lim; mid <<= 1)
{
complex I = complex{cos(Pi / mid), inv * sin(Pi / mid)};
for(re int i = 0; i < lim; i += (mid << 1))
{
complex W = complex{1, 0};
for(re int j = 0; j < mid; j++, W = W * I)
{
complex x = a[i + j], y = W * a[i + j + mid];
a[i + j] = x + y, a[i + j + mid] = x - y;
}
}
}
}
int main()
{
n = read(), m = read();
for(re int i = 0; i <= n; i++) a[i].x = read();
for(re int i = 0; i <= m; i++) b[i].x = read();
int limit = 1;
while (limit <= n + m) limit <<= 1;
int bit = 0;
while ((1 << bit) < limit) ++bit;
for(re int i = 0; i < limit; i++) rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << (bit - 1));
FFT(a, limit, 1), FFT(b, limit, 1);
for(re int i = 0; i < limit; i++) a[i] = a[i] * b[i];
FFT(a, limit, -1);
for(re int i = 0; i <= n + m; i++) printf("%d ", (int)(a[i].x / limit + 0.5));
}
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