快速傅里叶变换(FFT)，就是在O(nlogn)的时间内求出多项式系数矩阵a = (a1, a2, a3...an)的离散傅里叶变换(DFT)矩阵y = (y1, y2...yn)
如果你对傅里叶变换不是很了解，不要去网上查博客，打开算法导论的第30章，认认真真读一遍就懂了！！！！！！！！！
这是以n=4为例手推的FFT过程
#include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include using namespace std; const double eps(1e-8); typedef long long lint; const double PI = acos(-1.0); struct Complex { double real, image; Complex(double _real, double _image) { real = _real; image = _image; } Complex(){} }; Complex operator + (const Complex &c1, const Complex &c2) { return Complex(c1.real + c2.real, c1.image + c2.image); } Complex operator - (const Complex &c1, const Complex &c2) { return Complex(c1.real - c2.real, c1.image - c2.image); } Complex operator * (const Complex &c1, const Complex &c2) { return Complex(c1.real*c2.real - c1.image*c2.image, c1.real*c2.image + c1.image*c2.real); } int rev(int id, int len) { int ret = 0; for(int i = 0; (1 << i) < len; i++) { ret <<= 1; if(id & (1 << i)) ret |= 1; } return ret; } Complex A[140000]; void FFT(Complex* a, int len, int DFT)//对a进行DFT或者逆DFT, 结果存在a当中 //对应算法导论p537 { //Complex* A = new Complex[len]; 这么写会爆栈 for(int i = 0; i < len; i++) A[rev(i, len)] = a[i]; //下面说的二叉树是算法导论p537最上边的那颗二叉树 for(int s = 1; (1 << s) <= len; s++) //树有几层 { int m = (1 << s); Complex wm = Complex(cos(DFT*2*PI/m), sin(DFT*2*PI/m)); for(int k = 0; k < len; k += m) //每层几个结点 { Complex w = Complex(1, 0); for(int j = 0; j < (m >> 1); j++) //每个结点进行几次蝴蝶操作 { //蝴蝶操作 Complex t = w*A[k + j + (m >> 1)]; Complex u = A[k + j]; A[k + j] = u + t; A[k + j + (m >> 1)] = u - t; w = w*wm; } } } if(DFT == -1) for(int i = 0; i < len; i++) A[i].real /= len, A[i].image /= len; for(int i = 0; i < len; i++) a[i] = A[i]; return; } char numA[50010], numB[50010];//以每一位为系数, 那么多项式长度不超过50000 Complex a[140000], b[140000];//对应的乘积的长度不会超过100000, 也就是不超过(1 << 17) = 131072 int ans[140000]; int main() { while(~scanf("%s", numA)) { int lenA = strlen(numA); int sa = 0; while((1 << sa) < lenA) sa++; scanf("%s", numB); int lenB = strlen(numB); int sb = 0; while((1 << sb) < lenB) sb++; //那么乘积多项式的次数不会超过(1 << (max(sa, sb) + 1)) int len = (1 << (max(sa, sb) + 1)); for(int i = 0; i < len; i++) { if(i < lenA) a[i] = Complex(numA[lenA - i - 1] - '0', 0); else a[i] = Complex(0, 0); if(i < lenB) b[i] = Complex(numB[lenB - i - 1] - '0', 0); else b[i] = Complex(0, 0); } FFT(a, len, 1); FFT(b, len, 1);//把A和B换成点值表达 for(int i = 0; i < len; i++)//做点值表达的成乘法 a[i] = a[i]*b[i]; FFT(a, len, -1);//逆DFT换回原来的系数, 虚部一定是0 for(int i = 0; i < len; i++) ans[i] = (int)(a[i].real + 0.5);//取整误差的处理 for(int i = 0; i < len - 1; i++)//进位问题 { ans[i + 1] += ans[i] / 10; ans[i] %= 10; } bool flag = 0; for(int i = len - 1; i >= 0; i--)//注意输出格式的调整即可 { if(ans[i]) printf("%d", ans[i]), flag = 1; else if(flag || i == 0) printf("0"); } putchar('\n'); } return 0; }
