bzoj2194
http://www.lydsy.com/JudgeOnline/problem.php?id=2194
卷积。。。
卷积并不高深,其实卷积就是两个多项式相乘的系数,但是得满足一点条件,就是f[n]=a[i]*b[n-i],就是下标和固定。。。然后这道题下标和不固定,但是我们把b反过来,就是一个卷积了。每次和是固定的
但是输出的时候得输出从n-2n,因为c[n+k]=a[i]*b[n+k-i],n<=n+k<=2*n
#include<bits/stdc++.h> using namespace std; #define pi acos(-1) const int N = 600010; complex<double> a[N], b[N]; int n, m, l; int r[N]; void fft(complex<double> *a, int f) { for(int i = 0; i <= n; ++i) if(i < r[i]) swap(a[i], a[r[i]]); for(int i = 1; i < n; i <<= 1) { complex<double> w(cos(pi / i), f * sin(pi / i)); for(int p = i << 1, j = 0; j <= n; j += p) { complex<double> t(1, 0); for(int k = 0; k < i; ++k, t = t * w) { complex<double> x = a[j + k], y = t * a[j + k + i]; a[j + k] = x + y; a[j + k + i] = x - y; } } } } int main() { scanf("%d", &n); --n; for(int i = 0; i <= n; ++i) { int x, y; scanf("%d%d", &x, &y); a[i] = x; b[n - i] = y; } m = 2 * n; for(n = 1; n <= m; n <<= 1) ++l; for(int i = 0; i <= n; ++i) r[i] = (r[i >> 1] >> 1) | ((i & 1) << (l - 1)); fft(a, 1); fft(b, 1); for(int i = 0; i <= n; ++i) a[i] = a[i] * b[i]; fft(a, -1); for(int i = m / 2; i <= m; ++i) printf("%d\n", (int)(a[i].real() / n + 0.5)); return 0; }