HDU 1402 FFT 求 大数乘法 (四)

2014-11-23 21:12:45 · 作者: · 浏览: 7
nt j = revv(i, bits);
if (i < j)
swap(a[i], a[j]), swap(b[i], b[j]);
}
for (int len = 2; len <= n; len <<= 1)
{
int half = len >> 1;
double wmx = cos(2 * PI / len), wmy = sin(2 * PI / len);
if (rev) wmy = -wmy;
for (int i = 0; i < n; i += len)
{
double wx = 1, wy = 0;
for (int j = 0; j < half; j++)
{
double cx = a[i + j], cy = b[i + j];
double dx = a[i + j + half], dy = b[i + j + half];
double ex = dx * wx - dy * wy, ey = dx * wy + dy * wx;
a[i + j] = cx + ex, b[i + j] = cy + ey;
a[i + j + half] = cx - ex, b[i + j + half] = cy - ey;
double wnx = wx * wmx - wy * wmy, wny = wx * wmy + wy * wmx;
wx = wnx, wy = wny;
}
}
}
if (rev)
{
for (int i = 0; i < n; i++)
a[i] /= n, b[i] /= n;
}
}
int solve(int a[],int na,int b[],int nb,int ans[]) //两个数组求卷积,有时ans数组要开成long long
{
int len = max(na, nb), ln;
for(ln=0; L(ln) len=L(++ln);
for (int i = 0; i < len ; ++i)
{
if (i >= na) ax[i] = 0, ay[i] =0;
else ax[i] = a[i], ay[i] = 0;
}
fft(ax, ay, len, 0);
for (int i = 0; i < len; ++i)
{
if (i >= nb) bx[i] = 0, by[i] = 0;
else bx[i] = b[i], by[i] = 0;
}
fft(bx, by, len, 0);
for (int i = 0; i < len; ++i)
{
double cx = ax[i] * bx[i] - ay[i] * by[i];
double cy = ax[i] * by[i] + ay[i] * bx[i];
ax[i] = cx, ay[i] = cy;
}
fft(ax, ay, len, 1);
for (int i = 0; i < len; ++i)
ans[i] = (int)(ax[i] + 0.5);
return len;
}
int solve(long long a[], int na, int ans[]) //自己跟自己求卷积,有时候ans数组要开成long long
{
int len = na, ln;
for(ln = 0; L(ln) < na; ++ln);
len=L(++ln);
for(int i = 0; i < len; ++i)
{
if (i >= na) ax[i] = 0, ay[i] = 0;
else ax[i] = a[i], ay[i] = 0;
}
fft(ax, ay, len, 0);
for(int i=0; i {
double cx = ax[i] * ax[i] - ay[i] * ay[i];
double cy = 2 * ax[i] * ay[i];
ax[i] = cx, ay[i] = cy;
}
fft(ax, ay, len, 1);

for(int i=0; i ans[i] = ax[i] + 0.5;
return len;
}