FFT-Fast Fourier Transformation

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Honestly, I don't want to write this thing at all ......

But delayyy has written all of them. Can you tell me if I can?> <~

For more information, see:

1. algorithm collection-polynomial multiplication (Zhang Jialin)

2. Introduction to algorithms: chapter on Fast Fourier Transformation.

3. The method for getting an integer from point X in All Integer FFT (Baidu space aekdycoin)

　

　　

I wrote the proof of the plural ...... After a long time.

Set W (k, n) to the unit root of 1 n times(Note that in the case of a plural number, there are exactly n solutions). W (k, n) has the following properties:

W (k, n) =Cos (2 * π * k/n) + I * sin (2 * π * k/N)

Half discount:W (k, n) = W (2 k, 2n) (directly approx)

Periodicity:W (k, n) = W (K + n, n) (the period of the trigonometric function is N)

Semi-inverse:W (k, n) =-W (K + n/2, n) (same as above, you can understand YY trigonometric functions)

　

The formula sub-Proofs in the FFT main algorithm are as follows:

S (x) = g (x) + x · f (x)

S (w (k, n) = g (w (k, n) ²) + W (k, n) · F (w (k, n) ²)

S (w (k, n) = g (w (k, n/2) + W (k, n) · F (w (k, n/2 ))

And:

S (w (K + n/2, n) = g (w (k, n/2) + W (K + n/2, n) · F (w (k, n/2 ))

S (w (K + n/2, n) = g (w (k, n/2)-W (K + n/2, n) · F (w (k, n/2 ))

　

In addition, the proof of interpolation after the main algorithm will not be put ...... Please Google

The above proof is that the division of FFT is still very powerful ......

　

Refer:Csdn edevil live FFT high-precision Multiplication

Csdn delayyy FFT

　

Below is a very frustrating code (Using FFT to optimize high-precision multiplication)

# Include <cmath> # include <cstdio> # include <cstdlib> # include <algorithm> const double Pi = ACOs (double)-1); Using namespace STD; char s [10010]; long C [10010]; int len_a, len_ B, N, T; struct line {Double X, Y;} A [10010], B [10010], W [10010], Y [10010]; line operator-(Line & A, line & B) {line C; C. X =. x-B. x; C. y =. y-B. y; return C;} line operator + (line & A, line & B) {line C; C. X =. X + B. x; c. Y =. Y + B. y; return C;} line operator * (line & A, line & B) {line C; C. X =. x * B. x-. y * B. y; C. y =. x * B. Y +. y * B. x; return C;} void FFT (line * a, int S, int t) {int Len = n> T; If (LEN = 1) return; line wk; FFT (A, s, t + 1); FFT (A, S + (1 <t), t + 1); For (int K = 0; k <(LEN> 1); ++ K) {int P = S + (k <(t + 1 )); wk = W [k <t] * A [p + (1 <t)]; y [k] = A [p] + wk, Y [K + (Len> 1)] = A [p]-wk;} For (int K = 0; k <Len; ++ k) A [S + (k <t)] = Y [k];} void set (char * s, line * a, Int & Len) // read, four digits one press {int Lenk, STA, I; gets (S + 1); Lenk = strlen (S + 1); STA = Lenk % 4; Len = (Lenk + 4)/4-1; for (I = 1; I <= sta; ++ I) A [Len]. X = A [Len]. x * 10 + s [I]-'0'; For (; I <= Lenk; I + = 4) A [-- Len]. X = (s [I]-'0') * 1000 + (s [I + 1]-'0 ') * 100 + (s [I + 2]-'0') * 10 + s [I + 3]-'0'; Len = (Lenk + 3)/4;} int main () {freopen ("input.txt", "r", stdin ); freopen ("output.txt", "W", stdout); set (s, A, len_a); Set (S, B, len_ B); n = len_a + len_ B; for (t = 1; t <n; t <= 1); n = t; for (INT I = 0; I <n; ++ I) W [I]. X = cos (pI * I * 2/N), W [I]. y = sin (pI * 2 * I/n); FFT (A, 0, 0); // FFT (B, 0, 0) of the point value process ); // B-based point value process for (INT I = 0; I <n; ++ I) A [I] = A [I] * B [I], W [I]. y = -W [I]. y; // inverse matrix processing FFT (A, 0, 0); // Interpolation Process for (INT I = 0; I <n; ++ I) c [I] + = (long) round (A [I]. x/N), C [I + 1] = C [I]/10000, C [I] %= 10000; int Top = N; while (! C [Top] & top> 0) -- top; printf ("% d", C [Top]); For (INT I = Top-1; I> = 0; -- I) printf ("% 04d", C [I]);}