#98. 【模板】多项式乘法

https://sjzezoj.com/submission/65266

S2OJ/98/98.cpp

@@ -0,0 +1,72 @@
+#include <iostream>
+#include <algorithm>
+#include <cmath>
+#include <complex>
+#include <vector>
+
+using std::cin;
+using std::cout;
+const char endl = '\n';
+
+const double PI = std::acos(-1);
+
+void FFT(std::vector<std::complex<double>>& a) {
+    if (a.size() == 1) return;
+
+    int m = a.size() >> 1;
+    std::vector<std::complex<double>> a0, a1;
+
+    for (int i = 0; i < m; i++) {
+        a0.emplace_back(a[i << 1]);
+        a1.emplace_back(a[i << 1 | 1]);
+    }
+
+    FFT(a0), FFT(a1);
+
+    std::complex<double>
+        w0{std::cos(PI / m), std::sin(PI / m)},
+        w1{1.0, 0.0};
+
+    for (int i = 0; i < m; i++) {
+        a[i] = a0[i] + w1 * a1[i];
+        a[i + m] = a0[i] - w1 * a1[i];
+        w1 *= w0;
+    }
+}
+
+int main() {
+    std::ios::sync_with_stdio(false);
+    cin.tie(nullptr);
+
+    int n, m;
+
+    cin >> n >> m;
+
+    int k = 1 << (std::__lg(n + m) + 1);
+    std::vector<std::complex<double>> f(k), g(k);
+
+    for (int i = 0; i <= n; i++) {
+        cin >> f[i];
+    }
+
+    for (int i = 0; i <= m; i++) {
+        cin >> g[i];
+    }
+
+    FFT(f), FFT(g);
+
+    for (int i = 0; i < k; i++) {
+        f[i] *= g[i];
+    }
+
+    FFT(f);
+    std::reverse(f.begin() + 1, f.end());
+
+    for (int i = 0; i <= n + m; i++) {
+        cout << static_cast<int>(std::round(f[i].real() / k)) << ' ';
+    }
+
+    cout << endl;
+
+    return 0;
+}