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HDU1402 FFT高精度乘法模板題

#include<bits/stdc++.h>
using namespace std;
//HDU 1402 求高精度乘法
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 必須為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 = 50005*2; Complex x1[MAXN],x2[MAXN]; char str1[MAXN/2],str2[MAXN/2]; int sum[MAXN]; int main() { while(scanf("%s%s",str1,str2)==2) { int len1 = strlen(str1); int len2 = strlen(str2); int len = 1; while(len < len1*2 || len < len2*2)len<<=1; for(int i = 0; i < len1; i++) x1[i] = Complex(str1[len1-1-i]-'0',0); for(int i = len1; i < len; i++) x1[i] = Complex(0,0); for(int i = 0; i < len2; i++) x2[i] = Complex(str2[len2-1-i]-'0',0); for(int i = len2; 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 < len; i++) { sum[i+1]+=sum[i]/10; sum[i]%=10; } len = len1+len2-1; while(sum[len] <= 0 && len > 0)len--; for(int i = len; i >= 0; i--) printf("%c",sum[i]+'0'); printf("\n"); } return 0; }