洛谷OJ P3803 【模板】多項式乘法(FFT)
阿新 • • 發佈:2018-12-15
題目思路:FFT模板題
AC程式碼:
// luogu-judger-enable-o2 #include <stdio.h> #include <string.h> #include <iostream> #include <algorithm> #include <math.h> using namespace std; const double PI = acos(-1.0); //複數結構體 struct Complex{ double x,y; //實部和虛部, x+yi Complex(double _x=0.0,double _y=0.0){ x=_x,y=_y; } Complex operator - (const Complex &b)const{ return Complex(x-b.x,y-b.y); } Complex operator + (const Complex &b)const{ return Complex(x+b.x,y+b.y); } Complex operator * (const Complex &b)const{ return Complex(x*b.x-y*b.y,x*b.y+y*b.x); } }; /* * 進行FFT和IFFT前的反轉變換。 * 位置i和 (i二進位制反轉後位置)互換 * len必須去2的冪 */ void change(Complex y[],int len) { int i,j,k; for(i = 1, j = len/2;i < len-1; i++) { if(i < j)swap(y[i],y[j]); //交換互為小標反轉的元素,i<j保證交換一次 //i做正常的+1,j左反轉型別的+1,始終保持i和j是反轉的 k = len/2; while( j >= k) { j -= k; k /= 2; } if(j < k) j += k; } } /* * 做FFT * len必須為2^k形式, * on==1時是DFT,on==-1時是IDFT */ void fft(Complex y[],int len,int on) { change(y,len); for(int h = 2; h <= len; h <<= 1) { Complex wn(cos(-on*2*PI/h),sin(-on*2*PI/h)); for(int j = 0;j < len;j+=h) { Complex w(1,0); for(int k = j;k < j+h/2;k++) { Complex u = y[k]; Complex t = w*y[k+h/2]; y[k] = u+t; y[k+h/2] = u-t; w = w*wn; } } } if(on == -1) for(int i = 0;i < len;i++) y[i].x /= len; } const int MAXN = 4000010; Complex x1[MAXN],x2[MAXN]; char str1[MAXN/2],str2[MAXN/2]; int sum[MAXN]; int main() { int n,m,x; while(scanf("%d%d",&n,&m)!=EOF) { int len = 1; while(len < n*2 || len < m*2)len<<=1; for(int i = 0;i < n+1;i++){ scanf("%d",&x); x1[i] = Complex(x,0); } for(int i = n+1;i < len;i++) x1[i] = Complex(0,0); for(int i = 0;i < m+1;i++){ scanf("%d",&x); x2[i] = Complex(x,0); } for(int i = m+1;i < len;i++) x2[i] = Complex(0,0); //求DFT 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++) sum[i] = (int)(x1[i].x+0.5); for(int i=0;i<n+m+1;i++) printf("%d%c",sum[i],(i==n+m)?'\n':' '); } return 0; }