快速傅里叶变换计算大整数乘法 code

快速傅里叶变换计算大整数乘法 code

2011-02-28 19:10:00

// a->A, b->B C->c  用三次快速傅立叶变换。

#include <stdio.h>
#include <string.h>
#include <math.h>

#define N 50009

char s[N];
int La,Lb,a[N+N],b[N+N];

double pi = acos(-1.0);

struct Num {
	double a,b;
}
A[N+N],B[N+N],C[N+N];

Num operator+ (Num aa, Num bb) {
	Num ret;
	ret.a = aa.a+bb.a; ret.b = aa.b+bb.b;
	return ret;
}
Num operator- (Num aa, Num bb) {
	Num ret;
	ret.a = aa.a-bb.a; ret.b = aa.b-bb.b;
	return ret;
}
Num operator* (Num aa, Num bb) {
	Num ret;
	ret.a = aa.a*bb.a - aa.b*bb.b;
	ret.b = aa.a*bb.b + aa.b*bb.a;
	return ret;
}

Num W(int n, int k) {
	Num ret;
	ret.a = cos(-pi*k*2/n);
	ret.b = sin(-pi*k*2/n);
	return ret;
}

void DFT(int L, int R, Num from[], Num X[])
{
	if(L+1 == R)
	{
		X[L] = from[L];
		return;
	}

	int i,j,k;
	Num T;

	for(i=L;i<R;i++) X[i] = from[i];
	j = L; k = (L+R)/2;
	for(i=L;i<R;i+=2)
	{
		from[j++] = X[i]; from[k++] = X[i+1];
	}

	DFT(L, (L+R)/2, from, X);
	DFT((L+R)/2, R, from, X);

	for(i=L;i<(L+R)/2;i++)
	{
		T = X[i];
		X[i] = T + W(R-L, i-L)*X[i+(R-L)/2];
		X[i+(R-L)/2] = T - W(R-L, i-L)*X[i+(R-L)/2];
	}
}

int main()
{
	int i;
	while(scanf("%s",s) != EOF)
	{
		La = strlen(s);
		for(i=0;i<La;i++) a[i] = s[La-i-1]-48;
		scanf("%s",s);
		Lb = strlen(s);
		for(i=0;i<Lb;i++) b[i] = s[Lb-i-1]-48;

		i=1; while(i<La+Lb-1) i = i*2;
		for(;La<i;La++) a[La] = 0;
		for(;Lb<i;Lb++) b[Lb] = 0;

		for(i=0;i<La;i++) {
			A[i].a = a[i]; A[i].b = 0;
			B[i].a = b[i]; B[i].b = 0;
		}

		DFT(0, La, B, C);
		DFT(0, Lb, A, B);

		for(i=0;i<La;i++) B[i] = B[i]*C[i];
		DFT(0, La, B, C);

		C[La] = C[0]; b[0] = 0;
		for(i=1;i<=La;i++)
		b[i] = (int)(C[i].a/La + 0.5);

		for(i=La;i>0;i--)
		{
			b[i-1] += b[i]/10; b[i] %= 10;
		}

		i = 0; while(i < La && b[i] == 0) i++;
		for(;i<=La;i++) printf("%d",b[i]); printf("\n");
	}
	return 0;
}