A * B Problem Plus HDU - 1402 (FFT)
A * B Problem Plus
第一道FFT...
1 #include <iostream> 2 #include <complex> 3 #include <cstdio> 4 #include <cstring> 5 #include <cmath> 6 #include <algorithm> 7 using namespace std; 8 9 const int maxn = 200010; 10 const double pi = acos(-1.0); 11 complex<double> omega[maxn], inverse[maxn]; 12 13 void init(int n){ 14 for(int i = 0; i < n; i++){ 15 omega[i] = complex<double>(cos(2 * pi / n * i), sin(2 * pi / n * i)); 16 inverse[i] = conj(omega[i]); 17 } 18 } 19 void transform(complex<double> *a, int n, complex<double> *omega){ 20 int k = 0; 21 while((1 << k) < n) k++; 22 for(int i = 0; i < n; i++){ 23 int t = 0; 24 for(int j = 0; j < k; j++) if(i & (1 << j)) t |= (1 << (k - j - 1)); 25 if(i < t) swap(a[i], a[t]); 26 } 27 for(int l = 2; l <= n; l *= 2){ 28 int m = l / 2; 29 for(complex<double> *p = a; p != a + n; p += l){ 30 for(int i = 0; i < m; i++){ 31 complex<double> t = omega[n / l * i] * p[m + i]; 32 p[m + i] = p[i] - t; 33 p[i] += t; 34 } 35 } 36 } 37 } 38 void dft(complex<double> *a, int n){ 39 transform(a, n, omega); 40 } 41 void idft(complex<double> *a, int n){ 42 transform(a, n, inverse); 43 for(int i = 0; i < n; i++) a[i] /= n; 44 } 45 46 void mul(int *a, int n1, int *b, int n2, int *res){ 47 int n = 1; 48 while(n < n1 + n2) n <<= 1; 49 complex<double> c1[maxn], c2[maxn]; 50 for(int i = 0; i < n1; i++) c1[i].real(a[i]); 51 for(int i = 0; i < n2; i++) c2[i].real(b[i]); 52 init(n); 53 dft(c1, n); dft(c2, n); 54 for(int i = 0; i < n; i++) c1[i] *= c2[i]; 55 idft(c1, n); 56 for(int i = 0; i < n1 + n2 - 1; i++) res[i] = (int)(floor(c1[i].real() + 0.5)); 57 } 58 char s1[maxn], s2[maxn]; 59 int a[maxn], b[maxn], res[maxn]; 60 61 void in(){ 62 freopen("in.txt", "w", stdout); 63 for(int i= 0; i < 100; i++){ 64 cout<<rand() % 131313 <<endl; 65 } 66 } 67 int main(){ 68 // freopen("in.txt", "r", stdin); 69 // freopen("out.txt", "w", stdout); 70 while(scanf("%s %s", s1, s2) != EOF){ 71 memset(res, 0, sizeof res); 72 int n1 = strlen(s1), n2 = strlen(s2); 73 for(int i = 0; i < n1; i++) a[i] = s1[n1 - i - 1] - '0'; 74 for(int i = 0; i < n2; i++) b[i] =s2[n2 - i - 1] - '0'; 75 mul(a, n1, b, n2, res); 76 int m = n1 + n2 -1; 77 for(int i = 0; i < m; i++) res[i + 1] += res[i] / 10, res[i] %= 10; 78 if(res[m]) m++; 79 while(m >= 1 && !res[m - 1]) m--; 80 for(int i = m - 1; i >= 0; i--) printf("%d", res[i]); 81 if(m == 0) printf("0"); 82 puts(""); 83 } 84 }
