求Q(x)模X^n + 1的餘數多項式的FFT算法

模的餘數多項式，最高次數不超過n，令其餘數多項式的形式爲

 

1、因爲有n個解，對應餘數多項式有n個，分別是其中，0<=j<=n-1,而則是的n個解，而是容易求的，

,但是如果每個都往裏面帶入的話，會花費大量的計算量。

 

這裏提一種基於FFT的簡單運算，這裏要求。（這裏對爲什麼是基於FFT的我也不是很明白，有一說是

的離散傅里葉變換DFS，用快速傅里葉變換FFT計算得來的，DFS的公式這兒就不列出了）

 

2、計算算法，因爲，把按奇偶項分開，每次分開一半，直到最後只剩兩項的時候進行計算，然後一直遞歸向上，這樣就把大量的乘冪運算轉化爲了加法運算，下面給出C語言的代碼

 

//

#include "stdafx.h"

#define _CRT_SECURE_NO_WARNINGS
#include "stdlib.h"
#include "math.h"
#include "stdio.h"


#define K 3

#define Pi  3.1415927   //定義圓周率Pi
#define LEN sizeof(struct Compx)  //定義複數結構體大小

//-----定義複數結構體-----------------------
struct Compx
{
	double real;
	double imag;
}Compx;

//-----複數乘法運算函數---------------------
struct Compx mult(struct Compx b1, struct Compx b2)
{
	struct Compx b3;
	b3.real = b1.real*b2.real - b1.imag*b2.imag;
	b3.imag = b1.real*b2.imag + b1.imag*b2.real;
	return(b3);
}

//-----複數加法運算函數---------------------
struct Compx add(struct Compx a, struct Compx b)
{
	struct Compx c;
	c.real = a.real + b.real;
	c.imag = a.imag + b.imag;
	return(c);
}

struct Compx FFT(struct Compx *t , int n , struct Compx root , struct Compx result ) ;

int main(  ) 
{
	int N,i;
	N =8;

    int x[8];
    for( i = 0 ; i < N; i++ )
	{
    	scanf("%d",&x[i]);
	}
	
    struct  Compx * Source = (struct Compx *)malloc(N*LEN);			//爲結構體分配存儲空間
    struct  Compx * Result = (struct Compx *)malloc(N*LEN);
    struct  Compx * Root = (struct Compx *)malloc(N*LEN);

    
    //初始化======================================= 
    printf("\nSource初始化：\n");
	for (i = 0; i<N; i++)
	{
		Source[i].real = x[i];
		Source[i].imag = 0;
		printf("%.4f ", Source[i].real);
		printf("+%.4fj  ", Source[i].imag);
		printf("\n");
	}
	printf("\nResult初始化：\n");
	for (i = 0; i<N; i++)
	{
		Result[i].real = 0;
		Result[i].imag = 0;
		printf("%.4f ", Result[i].real);
		printf("+%.4fj  ", Result[i].imag);
		printf("\n");
	}
	printf("\nRoot初始化：\n");
	for (i = 0; i<N; i++)
	{
		Root[i].real = cos( Pi * ( 2 * i + 1) / (2 * N));
		Root[i].imag = sin( Pi * ( 2 * i + 1) / (2 * N));
		printf("%.4f ", Root[i].real);
		printf("+%.4fj  ", Root[i].imag);
		printf("\n");
	}
	

    for( i = 0 ; i < N ; i++) 
    {
    	Result[i] = FFT(Source , N , Root[i] , Result[i]);
	}
	
    //結果表示
	printf("\nResult計算結果：\n");
	for (i = 0; i<N; i++)
	{
		printf("%.4f ", Result[i].real);
		printf("+%.4fj  ", Result[i].imag);
		printf("\n");
	}
    
	return 0;
}

struct Compx FFT(struct Compx *t , int n , struct Compx root , struct Compx result ) 
{
	int i,j;
	struct  Compx * even = (struct Compx *)malloc((n / 2) * LEN);	
	struct  Compx * odd = (struct Compx *)malloc((n / 2) * LEN);
	
	//劃分奇偶項 
	for (i = 0 , j = 0 ; i < n; i += 2 , j++ )
	{
		even[j].real = t[i].real;
		even[j].imag = t[i].imag;
	}
	for (i = 1 , j = 0 ; i < n; i += 2 , j++)
	{
		odd[j].real = t[i].real;
		odd[j].imag = t[i].imag;
	}
	
	if(n == 2)
	{
		struct Compx s = add( even[0] , mult( root , odd[0]) );
		return add(result , s);
	}
	else
	{
		return add( FFT(even , n / 2 , mult(root , root) , result ) , mult( root , FFT(odd , n / 2 , mult(root , root) , result ) ) );
	}
}

歡迎指正！