【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

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#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：

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Description

给出n个数q_i，给出F_j的定义如下： 
 
令E_i=F_i/q_i，求E_i. 

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E_i。 
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<q_i<1000000000。 

Source

感谢nodgd放题