【HDU 1402】A * B Problem Plus（FFT）

最新推荐文章于 2021-10-21 20:07:10 发布

AC_alvin

最新推荐文章于 2021-10-21 20:07:10 发布

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本文链接：https://blog.youkuaiyun.com/u011332631/article/details/46662441

本文提供了一个快速傅里叶变换（FFT）的C++实现模板，用于解决多项式乘法问题。通过FFT进行高效的多项式乘法运算，代码包括复数定义、位逆置、递归FFT和逆FFT等关键部分。

这道题是快速傅里叶变换的最模板的题了，存个板子

#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
using namespace std;
const double pi = acos(-1.0);
const int N = 200005;
struct Complex{
    double r,i;
    Complex(double r = 0,double i = 0):r(r),i(i){}
    Complex operator + (const Complex &a)const{
        return Complex(r + a.r,i + a.i);
    }
    Complex operator - (const Complex &a)const{
        return Complex(r - a.r,i - a.i);
    }
    Complex operator * (const Complex &a)const{
        return Complex(r * a.r - i * a.i,r * a.i + i * a.r);
    }
    Complex operator / (const double &a)const{
        return Complex(r / a,i / a);
    }
};
char s[N];
Complex x1[N],x2[N],x3[N];
Complex tmp[N];
int res[N],tn;
int rev(int x){
    int res = 0;
    for(int i = 0 ; i < tn ; i ++){
        if(x & 1) res += 1 << tn - i - 1;
        x >>= 1;
    }
    return res;
}
//op == 1 DFT,op == -1 IDFT
void fft(Complex a[],int n,int op){
    for(int i = 0 ; i < n ; i ++) tmp[ rev(i) ] = a[i];
    for(int i = 0 ; i < n ; i ++) a[i] = tmp[i];
    for(int i = 1 ;  (1 << i) <= n ; i ++){
        int m = 1 << i;
        Complex wn(cos(op * 2 * pi / m),sin(op * 2 * pi / m));
        for(int k = 0 ; k < n ; k += m){
            Complex w(1,0),u,t;
            for(int j = 0 ; j < m / 2 ; j ++){
                u = a[k + j];
                t = w * a[k + j + m / 2];
                a[k + j] = u + t;
                a[k + j + m / 2] = u - t;
                w = w * wn;
            }
        }
    }
    if(op == -1) for(int i = 0 ; i < n ; i ++) a[i] = a[i] / n;
}
int main()
{
    int len1,len2,len;
    while(~scanf("%s",s)){
        len1 = strlen(s);
        for(int i = 0 ; i < len1 ; i ++)
            x1[len1 - 1 - i] = Complex(s[i] - '0',0);
        scanf("%s",s);
        len2 = strlen(s);
        for(int i = 0 ; i < len2 ; i ++)
            x2[len2 - 1 - i] = Complex(s[i] - '0',0);
        tn = ceil(log(len1 + len2 + 0.0) / log(2.0));
        len = 1 << tn;
        for(int i = len1 ; i < len ; i ++) x1[i] = Complex(0,0);
        for(int i = len2 ; i < len ; i ++) x2[i] = Complex(0,0);
        fft(x1,len,1);
        fft(x2,len,1);
        for(int i = 0 ; i < len ; i ++) x1[i] = x1[i] * x2[i];
        fft(x1,len,-1);
        for(int i = 0 ; i < len ; i ++) res[i] = int(x1[i].r + 0.5);
        for(int i = 0 ; i < len ; i ++){
            res[i + 1] += res[i] / 10;
            res[i] %= 10;
        }
        int tail = len - 1;
        while(tail && res[tail] == 0) tail --;
        while(tail >= 0) printf("%d",res[tail --]);
        printf("\n");
    }
    return 0;
}