# 【BZOJ 2194】快速傅立叶之二

#include<cmath>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int N = 400003;
const double Pi = acos(- 1.0);
int getint() {
int k = 0, fh = 1; char c = getchar();
for(; c < '0' || c > '9'; c = getchar())
if (c == '-') fh = -1;
for(; c >= '0' && c <= '9'; c = getchar())
k = k * 10 + c - '0';
return k * fh;
}
struct cp {
double r, i;
cp (double _r = 0.0, double _i = 0.0) : r(_r), i(_i) {}
cp operator + (const cp &x) {return cp(r + x.r, i + x.i);}
cp operator - (const cp &x) {return cp(r - x.r, i - x.i);}
cp operator * (const cp &x) {return cp(r * x.r - i * x.i, r * x.i + i * x.r);}
};
cp A[N], u, t;
int rev[N];
void DFT(cp *a, int n, int flag) {
for(int i = 0; i < n; ++i) A[rev[i]] = a[i];
for(int i = 0; i < n; ++i) a[i] = A[i];
for(int m = 2; m <= n; m <<= 1) {
cp wn(cos(2.0 * Pi / m * flag), sin(2.0 * Pi / m * flag));
int mid = m >> 1;
for(int i = 0; i < n; i += m) {
cp w(1.0);
for(int j = 0; j < mid; ++j) {
u = a[i + j], t = a[i + j + mid] * w;
a[i + j] = u + t;
a[i + j + mid] = u - t;
w = w * wn;
}
}
}
if (flag == -1)
for(int i = 0; i < n; ++i)
a[i].r /= n;
}
void init(int &n) {
int k = 1, ret, L = 0;
for(; k < n; k <<= 1, ++L);
n = k;
for(int i = 0; i < n; ++i) {
k = i; ret = 0;
for(int j = 0; j < L; ++j)
ret <<= 1, ret |= k & 1, k >>= 1;
rev[i] = ret;
}
}
void FFT(int *a, int *b, int *c, int la, int lb) {
static cp x[N], y[N];
int len = la + lb - 1;
init(len);
for(int i = 0; i < len; ++i)
x[i].r = a[i], x[i].i = 0;
for(int i = 0; i < len; ++i)
y[i].r = b[i], y[i].i = 0;
DFT(x, len, 1); DFT(y, len, 1);
for(int i = 0; i < len; ++i)
x[i] = x[i] * y[i];
DFT(x, len, -1);
for(int i = 0; i < len; ++i)
c[i] = (int) (x[i].r + 0.5);
}
int x[N], y[N], a[N], n;
int main() {
for(int i = 0; i < n; ++i)
for(int i = 0; i < n; ++i)
a[i] = x[n - i - 1];
FFT(y, a, x, n, n);
for(int i = 0; i < n; ++i)
a[i] = x[n - i - 1];
for(int i = 0; i < n; ++i)
printf("%d\n", a[i]);
return 0;
}


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posted @ 2016-04-25 20:10  abclzr  阅读(193)  评论(0编辑  收藏  举报