【BZOJ 3527】【ZJOI 2014】力
代换一下变成多项式卷积,这里是的答案是两个卷积相减,FFT求一下两个卷积就可以啦
详细的题解:http://www.cnblogs.com/iwtwiioi/p/4126284.html
#include<cmath> #include<cstdio> #include<cstring> #include<algorithm> using namespace std; const int N = 500003; const double Pi = acos(- 1.0); struct cp { double r, i; cp (double _r = 0.0, double _i = 0.0) : r(_r), i(_i) {} cp operator + (const cp &x) const {return cp(r + x.r, i + x.i);} cp operator - (const cp &x) const {return cp(r - x.r, i - x.i);} cp operator * (const cp &x) const {return cp(r * x.r - i * x.i, r * x.i + i * x.r);} }; int rev[N]; cp A[N]; void DFT(cp *a, int n, int flag) { for(int i = 0; i < n; ++i) A[rev[i]] = a[i]; for(int i = 0; i < n; ++i) a[i] = A[i]; for(int m = 2; m <= n; m <<= 1) { cp wn(cos(2.0 * Pi / m * flag), sin(2.0 * Pi / m * flag)); int mid = m >> 1; for(int i = 0; i < n; i += m) { cp w(1.0); for(int j = 0; j < mid; ++j) { cp u = a[i + j], t = a[i + j + mid] * w; a[i + j] = u + t; a[i + j + mid] = u - t; w = w * wn; } } } if (flag == -1) for(int i = 0; i < n; ++i) a[i].r /= n; } void init(int &n) { int k = 1, L = 0; for(; k < n; k <<= 1, ++L); n = k; for(int i = 0; i < n; ++i) { int t = i, ret = 0; for(int j = 0; j < L; ++j) ret <<= 1, ret |= (t & 1), t >>= 1; rev[i] = ret; } } void FFT(double *x, double *y, cp *a, cp *b, int len) { for(int i = 0; i < len; ++i) a[i].r = x[i], a[i].i = 0.0; for(int i = 0; i < len; ++i) b[i].r = y[i], b[i].i = 0.0; DFT(a, len, 1); DFT(b, len, 1); for(int i = 0; i < len; ++i) a[i] = a[i] * b[i]; DFT(a, len, -1); } cp a[N], b[N]; int n, len; double g[N], q[N], f[N], ans[N]; int main() { scanf("%d", &n); len = (n << 1) + 1; init(len); for(int i = 1; i <= n; ++i) scanf("%lf", &q[i]); for(int i = 1; i <= n; ++i) g[i] = 1.0 / i / i; for(int i = 0; i < n; ++i) f[i] = q[n - i]; FFT(q, g, a, b, len); for(int i = 1; i <= n; ++i) ans[i] = a[i].r; FFT(f, g, a, b, len); for(int i = 1; i <= n; ++i) ans[i] -= a[n - i].r; for(int i = 1; i <= n; ++i) printf("%.3lf\n", ans[i]); return 0; }
题面如下,BZOJ上没有题面喔:
Description
给出n个数qi,给出Fj的定义如下:
令Ei=Fi/qi,求Ei.
Input
第一行一个整数n。
接下来n行每行输入一个数,第i行表示qi。
Output
n行,第i行输出Ei。
与标准答案误差不超过1e-2即可。
Sample Input
5 4006373.885184 15375036.435759 1717456.469144 8514941.004912 1410681.345880
Sample Output
-16838672.693 3439.793 7509018.566 4595686.886 10903040.872
Hint
对于30%的数据,n≤1000。
对于50%的数据,n≤60000。
对于100%的数据,n≤100000,0<qi<1000000000。
Source
感谢nodgd放题
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