http://acm.hdu.edu.cn/showproblem.php?pid=1402
fft做O(nlog(n))大数乘法,kuangbin的模板
#include <stdio.h> #include <string.h> #include <iostream> #include <algorithm> #include <math.h> using namespace std; const double PI = acos(-1.0); //复数结构体 struct Complex { double x,y;//实部和虚部 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); } }; /* * 进行FFT和IFFT前的反转变换。 * 位置i和 (i二进制反转后位置)互换 * len必须去2的幂 */ 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]); //交换互为小标反转的元素,i<j保证交换一次 //i做正常的+1,j左反转类型的+1,始终保持i和j是反转的 k = len/2; while(j >= k) { j -= k; k /= 2; } if(j < k)j += k; } } /* * 做FFT * len必须为2^k形式, * on==1时是DFT,on==-1时是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 = 200010; Complex x1[MAXN],x2[MAXN]; char str1[MAXN/2],str2[MAXN/2]; int sum[MAXN]; int main() { while(~scanf("%s%s",str1,str2)) { 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("%d",sum[i]); printf("\n"); } return 0; }