HDU 1402 FFT 大数乘法

$A * B$

FFT模板题，找到了一个看起来很清爽的模板

 

/** @Date    : 2017-09-19 22:12:08
  * @FileName: HDU 1402 FFT 大整数乘法.cpp
  * @Platform: Windows
  * @Author  : Lweleth (SoungEarlf@gmail.com)
  * @Link    : https://github.com/
  * @Version : $Id$
  */
#include <bits/stdc++.h>
#define LL long long
#define PII pair
#define MP(x, y) make_pair((x),(y))
#define fi first
#define se second
#define PB(x) push_back((x))
#define MMG(x) memset((x), -1,sizeof(x))
#define MMF(x) memset((x),0,sizeof(x))
#define MMI(x) memset((x), INF, sizeof(x))
using namespace std;

const int INF = 0x3f3f3f3f;
const int N = 2e5+20;
const double eps = 1e-8;
const double pi = acos(-1.0);

struct Complex
{
	double a, b;
	Complex(){}
	Complex(double _a, double _b):a(_a),b(_b){}
	Complex(double _a):a(_a),b(0.0){}
	inline Complex operator +(const Complex &z)const{
		return Complex(a + z.a, b + z.b);
	}
	inline Complex operator -(const Complex &z)const{
		return Complex(a - z.a, b - z.b);
	}
	inline Complex operator *(const Complex &z)const{
		return Complex(a * z.a - b * z.b, a * z.b + b * z.a);
	}
	inline Complex operator /(const Complex &z)const{
		double m = z.a * z.a + z.b * z.b;
		return Complex((a * z.a + b * z.b) / m, (z.a * b - z.b * a) / m);
	}
}A[N], B[N];
int L;
int rev[N];
int ans[N];
char a[N], b[N];

void init()
{
	MMF(A);
	MMF(B);
	MMF(ans);
	L = 0;//?
}

int reverse(int x,int r)  //蝴蝶操作  
{  
    int ans=0;  
    for(int i=0; i<r; i++){
        if(x&(1<<i)){
            ans+=1<<(r-i-1);  
        }  
    }  
    return ans;  
}  

void rader(LL f[], int len)
{
	for(int i = 1, j = len >> 1; i < len - 1; i++)
	{
		if(i < j) swap(f[i], f[j]);
		int k = len >> 1;
		while(j >= k)
		{
			j -= k;
			k >>= 1;
		}
		if(j < k) j += k;
	}
}
void FFT(Complex c[], int nlen, int on)
{
	Complex wn, w, x, y;
	for(int i = 0; i < nlen; i++)
		if(i < rev[i]) swap(c[i], c[rev[i]]);
	for(int i = 1; i < nlen; i <<= 1)
	{
		wn = Complex(cos(pi/i), sin(pi/i) * on);
		for(int p = i << 1, j = 0; j < nlen; j+= p)
		{
			w = Complex(1, 0);
			for(int k = 0; k < i; k++, w = w * wn)
			{
				x = c[j + k];
				y = w * c[j + k + i];
				c[j + k] = x + y;
				c[j + k + i] = x - y;
			}
		}
	}
	if(on == -1)
		for(int i = 0; i < nlen; i++)
			c[i].a /= (double)nlen;
}

void work(Complex a[], Complex b[], int len)
{
	FFT(a, len, 1); 
	FFT(b, len, 1);
	for(int i = 0; i < len; i++)
		a[i] = a[i] * b[i];
	FFT(a, len, -1);
}
int main()
{
	while(~scanf("%s%s", a, b))
	{
		init();
		int alen = strlen(a);
		int blen = strlen(b);
		int len = alen + blen;
		//getlen
		int n;
		for(n = 1; n < len - 1; n <<= 1, L++);
		//rev 
		for(int i = 0; i < n; i++)
			rev[i] = (rev[i>>1] >> 1) | ((i & 1) << (L - 1));
		//deal
		for(int i = 0; i < alen; i++) 
			A[i] = a[alen - i - 1] - '0';
		for(int i = 0; i < blen; i++) 
			B[i] = b[blen - i - 1] - '0';
		work(A, B, n);
		///
		for(int i = 0; i <= len; i++)
			ans[i] = (int)(A[i].a + 0.5);
		for(int i = 0; i <= len; i++)
			ans[i + 1] += ans[i] / 10, ans[i] %= 10;
		while(ans[len] == 0 && len)
			len--;
		for(int i = len; ~i; i--)
			printf("%c", ans[i] + '0');
		printf("\n");
	}
    return 0;
}