[BZOJ2179]FFT快速傅立叶（FFT）

题目描述

传送门

题解

将乘法的每一位看成是多项式的系数 实际上就是两个多项式相乘 直接FFT 处理进位

代码

#include<algorithm> #include<iostream> #include<cstring> #include<cstdio> #include<cmath> using namespace std; #define N 200005 const double pi=acos(-1.0); int n,m,L,R[N],ans[N]; struct complex { double x,y; complex(double X=0,double Y=0) { x=X,y=Y; } }a[N],b[N]; complex operator + (complex a,complex b) {return complex(a.x+b.x,a.y+b.y);} complex operator - (complex a,complex b) {return complex(a.x-b.x,a.y-b.y);} complex operator * (complex a,complex b) {return complex(a.x*b.x-a.y*b.y,a.x*b.y+a.y*b.x);} void FFT(complex a[N],int opt) { for (int i=0;i<n;++i) if (i<R[i]) swap(a[i],a[R[i]]); for (int k=1;k<n;k<<=1) { complex wn=complex(cos(pi/k),opt*sin(pi/k)); for (int i=0;i<n;i+=(k<<1)) { complex w=complex(1,0); for (int j=0;j<k;++j,w=w*wn) { complex x=a[i+j],y=w*a[i+j+k]; a[i+j]=x+y,a[i+j+k]=x-y; } } } } int main() { scanf("%d",&n);--n; char ch=getchar(); for (int i=n;i>=0;--i) a[i].x=(double)(getchar()-'0'); ch=getchar(); for (int i=n;i>=0;--i) b[i].x=(double)(getchar()-'0'); m=n+n; for (n=1;n<=m;n<<=1) ++L; for (int i=0;i<n;++i) R[i]=(R[i>>1]>>1)|((i&1)<<(L-1)); FFT(a,1);FFT(b,1); for (int i=0;i<=n;++i) a[i]=a[i]*b[i]; FFT(a,-1); for (int i=0;i<=m;++i) ans[i]=(int)(a[i].x/n+0.5); for (int i=0;i<=m;++i) { ans[i+1]+=ans[i]/10; ans[i]%=10; } while (ans[m+1]) { ++m; ans[m+1]+=ans[m]/10; ans[m]%=10; } for (int i=m;i>=0;--i) putchar(ans[i]+'0'); puts(""); }