Description
给出序列{qi},求
Ei=∑j<iqj(i−j)2−∑j>iqj(i−j)2
Solution
考虑:
令 pn−i=qi
Ei=∑j=1i−1qjgi−j−∑j=i+1nqjgj−i=∑j=1i−1qjgi−j−∑j=0n−i−1pjgn−j−i
然后FFT求卷积即可。
Code
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef complex<double> cpx;
#define rep(i, j) for (register int i = 0, i##_end_ = (j); i < i##_end_; ++ i)
#define For(i , j , k) for (register int i = (j) , i##_end_ = (k) ; i <= i##_end_ ; ++ i)
#define Fordown(i , j , k) for (register int i = (j) , i##_end_ = (k) ; i >= i##_end_ ; -- i)
#define Set(a , b) memset(a , b , sizeof(a))
#define pb(a) push_back(a)
#define mp(a, b) make_pair(a, b)
#define ALL(a) (a).begin(), (a).end()
#define SZ(a) ((int)(a).size())
#define fir first
#define sec second
#define INF (0x3f3f3f3f)
#define INF1 (2139062143)
#define Mod (1000000007)
#ifdef hany01
#define debug(...) fprintf(stderr , __VA_ARGS__)
#else
#define debug(...)
#endif
template <typename T> inline bool chkmax(T &a , T b) { return a < b ? (a = b , 1) : 0; }
template <typename T> inline bool chkmin(T &a , T b) { return b < a ? (a = b , 1) : 0; }
inline void File()
{
#ifdef hany01
freopen("3527.in" , "r" , stdin);
freopen("3527.out" , "w" , stdout);
#endif
}
const double pi = acos(-1.0);
const int maxn = 1 << 21;
int n, m, cnt, rev[maxn];
cpx q[maxn], p[maxn], g[maxn];
inline void Init()
{
scanf("%d", &n);
For(i, 1, n) scanf("%lf", &q[i].real()), p[n - i].real() = q[i].real(), g[i].real() = 1.0 / (i * 1ll * i);
for (m = n << 1, n = 1; n < m; n <<= 1) ++ cnt;
For(i, 0, n) rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << (cnt - 1));
}
inline void FFT(cpx *a, int type)
{
For(i, 0, n) if (rev[i] > i) swap(a[i], a[rev[i]]);
for (register int i = 2; i <= n; i <<= 1)
{
register cpx wn(cos(2 * pi / i), sin(2 * pi / i) * type);
for (register int j = 0; j < n; j += i)
{
register cpx w(1, 0);
rep(k, i >> 1)
{
register cpx x = a[j + k], y = a[j + k + (i >> 1)] * w;
a[j + k] = x + y; a[j + k + (i >> 1)] = x - y;
w = w * wn;
}
}
}
}
inline void Solve()
{
FFT(q, 1); FFT(p, 1); FFT(g, 1);
For(i, 0, n) p[i] *= g[i], q[i] *= g[i];
FFT(q, -1); FFT(p, -1);
}
inline void Print()
{
For(i, 1, m >> 1) printf("%.6lf\n", (q[i].real() - p[(m >> 1) - i].real()) / n);
}
int main()
{
File();
Init();
Solve();
Print();
return 0;
}
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