多項式相關演算法模板

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多項式乘法

 1 #include <bits/stdc++.h> 2 using namespace std; 3 const double pi = acos(-1); 4 const int LEN = 1e5 + 5; 5 int n, m, N; 6 namespace ploy { 7     struct comp { 8         double x, y; 9     } a[LEN * 4], b[LEN * 4];10     comp operator + (const comp &x,const comp &y) {11         return (comp){x.x + y.x, x.y + y.y};12     }13     comp operator - (const comp &x,const comp &y) {14         return (comp){x.x - y.x, x.y - y.y};15     }16     comp operator * (const comp &x,const comp &y) {17         return (comp){x.x * y.x - x.y * y.y, x.x * y.y + x.y * y.x};18     }19     void FFT(comp *a, int n, int x) {20         for (int i = n >> 1, j = 1; j < n; j++) {21             if (i < j) swap(a[i], a[j]);22             int k = n >> 1;23             for (; k & i; i ^= k, k >>= 1);24             i ^= k;25         }26         for (int m = 2; m <= n; m <<= 1) {27             comp w = (comp){cos(2.0 * pi * x / m), sin(2.0 * pi * x / m)};28             for (int i = 0; i + m <= n; i += m) {29                 comp t = (comp){1, 0};30                 for (int j = i; j < i + (m >> 1); j++) {31                     comp u = a[j];32                     comp v = t * a[j + (m >> 1)];33                     a[j] = u + v;34                     a[j + (m >> 1)] = u - v;35                     t = t * w;36                 }37             }38         }39         if (x == -1) {40             for (int i = 0; i < n; i++) a[i].x /= n;41         }42     }43 }44 using namespace ploy;45 int main() {46     scanf("%d %d", &n, &m);47     n++, m++;48     N = 1;49     while (N < n + m - 1) N <<= 1;50     for (int i = 0; i < n; i++) scanf("%lf", &a[i].x);51     for (int i = 0; i < m; i++) scanf("%lf", &b[i].x);52     FFT(a, N, 1);53     FFT(b, N, 1);54     for (int i = 0; i < N; i++) a[i] = a[i] * b[i];55     FFT(a, N, -1);56     for (int i = 0; i < n + m - 2; i++) printf("%d ", (int)(a[i].x + 0.5));57     printf("%d\n", (int)(a[n + m - 2].x + 0.5));58     return 0;59 }

FFT

 

多項式相關演算法模板