# BZOJ 2179: FFT快速傅立叶

#include <cstdio>
#include <iostream>
#include <cmath>

#define Max 1000005
#define rg register
typedef double flo;
struct vec
{
flo x, y; vec (flo a = 0, flo b = 0) : x (a), y (b) { }

vec operator + (const vec &rhs)
{ return vec (x + rhs.x, y + rhs.y); }

vec operator - (const vec &rhs)
{ return vec (x - rhs.x, y - rhs.y); }

vec &operator /= (const flo &k)
{ return x /= k, y /= k, *this; }

vec operator * (const vec &rhs)
{ return vec (x * rhs.x - y * rhs.y, x * rhs.y + y * rhs.x); }

} a[Max], b[Max];
int rd[Max];
char s[Max];
const flo PI = acos (-1.0);
void DFT (vec *a, int N, int f = 1)
{
rg int i, j, k; vec wn, w, x, y;
for (i = 0; i < N; ++ i) if (rd[i] > i) std :: swap (a[i], a[rd[i]]);

for (k = 1; k < N; k <<= 1)
{
wn = vec (cos (PI / k), f * sin (PI / k));
for (j = 0; j < N; j += k << 1)
for (i = 0, w = vec (1, 0); i < k; ++ i, w = w * wn)
x = a[j + i], y = w * a[i + j + k], a[i + j] = x + y, a[i + j + k] = x - y;
}
if (f == -1) for (i = 0; i < N; ++ i) a[i] /= N;
}
void FFT (vec *a, vec *b, int N)
{
int p = N << 1, L = 0; rg int i;
for (N = 1; N <= p; N <<= 1) ++ L;
for (i = 0; i < N; ++ i)
rd[i] = (rd[i >> 1] >> 1) | ((i & 1) << (L - 1));
DFT (a, N), DFT (b, N);
for (i = 0; i <= N; ++ i) a[i] = a[i] * b[i];
DFT (a, N, -1);
}
int c[Max];
int main (int argc, char *argv[])
{
int N; scanf ("%d", &N); -- N; rg int i;
scanf ("%s", s);
for (i = 0; i <= N; ++ i) a[i].x = s[N - i] - '0';
scanf ("%s", s);
for (i = 0; i <= N; ++ i) b[i].x = s[N - i] - '0';
FFT (a, b, N);
int p = N * 2;
for (i = 0; i <= p; ++ i) c[i] = (int) (a[i].x + 0.5);
for (i = 0; i <= p; ++ i)
if (c[i] >= 10)
{
c[i + 1] += c[i] / 10, c[i] %= 10;
if (i == p) ++ p;
}
for (; c[p] == 0; -- p);
for (i = p; i >= 0; -- i) printf ("%d", c[i]);
putchar ('\n');
return 0;
}

myj 吊打我Orz，xxy 捆起来打我Orz，myl 文化课上天Orz， lrh 姿势水平敲高Orz， hkd 特别胖Orz%%%，cys 智商感人Orz，syl zz专业Orz，我没有学上， 我们未来一片光明
posted @ 2017-12-15 21:08  ZlycerQan  阅读(199)  评论(0编辑  收藏  举报