【FFT】 HDU 1402 A * B Problem Plus
通道：http://acm.hdu.edu.cn/showproblem.php?pid=1402
题意：计算A*B
代码：
1 #include <cstdio> 2 #include <cmath> 3 #include <cstring> 4 #include <algorithm> 5 6 using namespace std; 7 8 typedef long long ll; 9 10 const int MAX_N = 100007; 11 const int MAX_M = 100007; 12 const long double PI = acos(-1.0); 13 14 struct Complex { 15 long double r, i; 16 Complex(long double _r, long double _i) { 17 r = _r; 18 i = _i; 19 } 20 Complex operator + (const Complex &c) { 21 return Complex(c.r + r, c.i + i); 22 } 23 Complex operator - (const Complex &c) { 24 return Complex(r - c.r, i - c.i); 25 } 26 Complex operator * (const Complex &c) { 27 return Complex(c.r * r - c.i * i, c.r * i + c.i * r); 28 } 29 Complex operator / (const int &c) { 30 return Complex(r / c, i / c); 31 } 32 Complex(){} 33 }; 34 namespace FFT { 35 int rev(int id, int len) { 36 int ret = 0; 37 for(int i = 0; (1 << i) < len; ++i) { 38 ret <<= 1; 39 if(id & (1 << i)) ret |= 1; 40 } 41 return ret; 42 } 43 Complex A[MAX_M << 3]; 44 void FFT(Complex *a, int len, int DFT) { 45 for(int i = 0; i < len; ++i) A[rev(i, len)] = a[i]; 46 for(int s = 1; (1 << s) <= len; ++s) { 47 int m = (1 << s); 48 Complex wm = Complex(cos(PI * DFT * 2 / m), sin(PI * DFT * 2 / m)); 49 for(int k = 0; k < len; k += m) { 50 Complex w = Complex(1, 0); 51 for(int j = 0; j < (m >> 1); j++) { 52 Complex t = w * A[k + j + (m >> 1)]; 53 Complex u = A[k + j]; 54 A[k + j] = u + t; 55 A[k + j + (m >> 1)] = u - t; 56 w = w * wm; 57 } 58 } 59 } 60 if(DFT == -1) for(int i = 0; i < len; ++i) A[i] = A[i] / len; 61 for(int i = 0; i < len; i++) a[i] = A[i]; 62 } 63 }; 64 65 char a[MAX_M], b[MAX_M]; 66 Complex X[MAX_M << 2], Y[MAX_M << 2]; 67 int res[MAX_M << 2]; 68 69 int main() { 70 while (2 == scanf("%s%s", a, b)) { 71 int la = strlen(a), lb = strlen(b); 72 for (int i = 0; i < la; ++i) X[i] = Complex(a[la - i - 1] - '0', 0); 73 for (int i = 0; i < lb; ++i) Y[i] = Complex(b[lb - i - 1] - '0', 0); 74 int len = 1; 75 while (len < la || len < lb) len <<= 1; len <<= 1; 76 for (int i = la; i < len; ++i) X[i] = Complex(0, 0); 77 for (int i = lb; i < len; ++i) Y[i] = Complex(0, 0); 78 FFT::FFT(X, len, 1); FFT::FFT(Y, len, 1); 79 for (int i = 0; i < len; ++i) X[i] = X[i] * Y[i]; 80 FFT::FFT(X, len, -1); 81 for (int i = 0; i < len; ++i) res[i] = (int)(X[i].r + 0.5); 82 for (int i = 0; i < len; ++i) { 83 res[i + 1] += res[i] / 10; 84 res[i] %= 10; 85 } 86 len = la + lb - 1; 87 while (res[len] <= 0 && len > 0) --len; 88 for (int i = len; i >= 0; --i) printf("%d", res[i]); puts(""); 89 } 90 return 0; 91 }
