标签:else turn 我的朋友 要求 alt ini 大于 题解 统计
\(1 \le l \le n \le 10^5\)
写完之后觉得我死了,这应该是我做过的最复杂的分治NTT。
设\(p[i]\)表示第\(i\)个点概率,为了表达方便,先把\(p[i]\) reverse一下。
设\(P(i)=p[i]·x+1-p[i]\)
考虑设\(f[i]\)表示从第\(i\)个点出发的期望步数。
\(i<l,f[i]=0\)
设\(H_i(x)=\prod_{j=i-l+1}^{i}P(j)\)
\(i\ge l,f[i]=1+\sum_{j=0}^l f[i-j]*H_i[x^j]\)
\(j=0\)时,\(f[i-j]=f[i]\),把这一项移到等式右边,解个方程可得:
\(i \ge l,f[i]=(1+\sum_{j=1}^l f[i-j]*H_i[x^j])/(1-H_i[x^0])\)
得到了\(O(n^2)\)的做法。
设\(F_i=\sum_{j=1}^i f[j]*x^j\)
则\(f[i]=F_{i-1}*H_i[x^i]\)
考虑分治求\(f[]\),现在已经求出了\(f[1..x-1]\),要求\(f[x..y]\)
在这个分治的过程中,动态的维护两个函数\(F[x][y],G[x][y]\)
\(G[x][y]=\prod_{j=y-l+1}^{x}P(j)\)
它的含义即为\(f[x,y]\)的转移里都会用到\(P\)的乘积。
注意到\(y-l+1\)可能会大于\(x\),此时我们把意义推广,设\(SP\)为\(P\)的前缀积,则:
\(G[x][y]=\prod_{j=y-l+1}^{x}P(j)=SP[x]/SP[y-l]\)
\(F[x][y]=F_{x-1}*G[x][y]\)
如果能一直维护这两个函数,那么当\(x=y\)时直接取\(F[x][y]\)第\(x\)项的系数就好了。
设\(m=(x+y)/2\)
1.由\([x,y]\)递归至\([x,m]\)
\(G[x][m]=G[x][y]*\prod_{i=m}^{r-1}P(i-l+1)=G[x][y]*\prod_{i=m+1}^{r}P(i-l)\)
\(F[x][m]=F[x][y]*\prod_{i=m+1}^{r}P(i-l)\)
2.由\([x,y]\)递归至\([m+1,y]\)
\(G[m+1][y]=G[x][y]*\prod_{i=x+1}^{m+1}P(i)=G[x][y]*\prod_{i=x}^{m}P(i+1)\)
\(F[m+1][y]=F[x][y]*\prod_{i=x}^{m}P(i+1)+(F_m-F_{x-1})*G[m+1][y]\)
发现每次乘上的\(\prod P\)都可以通过分治NTT预处理出来
当然这么做并没有使复杂度降低。
观察\(F[x][y],G[x][y]\),在递归的过程中一直乘\(\prod P\),也就是次数最多+\(y-x\)。
最后要统计答案取的那一项的次数也在\([x,y]\)里。
那么只维护\({F[x][y]\over x^{max(0,x-(y-x))}}\)的前\(2(y-x)\)项和\(G[x][y]\)的前\(y-x\)项就可以了。
因为一切多项式的长度都是\(O(y-x)\),所以时间复杂度是\(O(n~log^2~n)\)
最后一个问题就是\(F[1][n]\)和\(G[1][n]\)怎么求?
\(F[1][n]\)显然是0.
\(G[1][n]=SP[1]/SP[n-l]\)
分治NTT之后再多项式求逆就可以求出\(G[1][n]\)。
#include<bits/stdc++.h>
#define fo(i, x, y) for(int i = x, _b = y; i <= _b; i ++)
#define ff(i, x, y) for(int i = x, _b = y; i < _b; i ++)
#define fd(i, x, y) for(int i = x, _b = y; i >= _b; i --)
#define ll long long
#define pp printf
#define hh pp("\n")
using namespace std;
const int mo = 998244353;
ll ksm(ll x, ll y) {
ll s = 1;
for(; y; y /= 2, x = x * x % mo)
if(y & 1) s = s * x % mo;
return s;
}
#define V vector<ll>
#define re resize
#define si size()
const int nm = 1 << 18;
namespace ntt {
ll w[nm], a[nm], b[nm]; int r[nm];
void build() {
for(int i = 1; i < nm; i *= 2) {
ll v = ksm(3, (mo - 1) / 2 / i);
w[i] = 1; ff(j, 1, i) w[i + j] = w[i + j - 1] * v % mo;
}
}
void dft(ll *a, int n, int f) {
ff(i, 0, n) {
r[i] = r[i / 2] / 2 + (i & 1) * (n / 2);
if(i < r[i]) swap(a[i], a[r[i]]);
} ll b;
for(int i = 1; i < n; i *= 2) for(int j = 0; j < n; j += 2 * i)
ff(k, 0, i) b = a[i + j + k] * w[i + k], a[i + j + k] = (a[j + k] - b) % mo, a[j + k] = (a[j + k] + b) % mo;
if(f == -1) {
reverse(a + 1, a + n);
b = ksm(n, mo - 2);
ff(i, 0, n) a[i] = (a[i] + mo) * b % mo;
}
}
void fft(V &p, V &q) {
int p0 = p.si + q.si - 1, n = 1;
for(; n < p0; n *= 2);
ff(i, 0, n) a[i] = b[i] = 0;
ff(i, 0, p.si) a[i] = p[i];
ff(i, 0, q.si) b[i] = q[i];
dft(a, n, 1); dft(b, n, 1);
ff(i, 0, n) a[i] = a[i] * b[i] % mo;
dft(a, n, -1);
p.re(p0);
ff(i, 0, p0) p[i] = a[i];
}
}
V operator * (V p, V q) {
ntt :: fft(p, q);
return p;
}
void dft(V &p, int f) {
ff(i, 0, p.si) ntt :: a[i] = p[i];
ntt :: dft(ntt :: a, p.si, f);
ff(i, 0, p.si) p[i] = ntt :: a[i];
}
V qni(V a) {
int a0 = a.si, n0 = 1;
while(n0 < a0) n0 *= 2;
V b; b.re(1); b[0] = ksm(a[0], mo - 2);
for(int n = 2; n <= n0; n *= 2) {
V d = b; d.re(n); b.re(2 * n);
V c = a; c.re(n); c.re(2 * n);
dft(c, 1); dft(b, 1);
ff(i, 0, b.si) b[i] = c[i] * b[i] % mo * b[i] % mo;
dft(b, -1); b.re(n);
ff(i, 0, b.si) b[i] = (2 * d[i] - b[i] + mo) % mo;
}
b.re(a0);
return b;
}
V operator + (V p, V q) {
p.re(max(p.si, q.si));
ff(i, 0, q.si) p[i] = (p[i] + q[i]) % mo;
return p;
}
const int N = 1e5 + 5;
int n, l;
ll x, y, p[N];
ll sp[N], np[N];
void Init() {
scanf("%d %d", &n, &l);
fo(i, 1, n) {
scanf("%lld %lld", &x, &y);
p[i] = x * ksm(y, mo - 2) % mo;
}
reverse(p + 1, p + n + 1);
sp[0] = 1; fo(i, 1, n) sp[i] = sp[i - 1] * (1 - p[i]) % mo;
np[n] = ksm(sp[n], mo - 2); fd(i, n, 1) np[i - 1] = np[i] * (1 - p[i]) % mo;
}
ll calc(int x) {
return sp[x] * np[x - l] % mo;
}
V t1[N * 4], t2[N * 4];
#define i0 i + i
#define i1 i + i + 1
void dg(int i, int x, int y) {
if(x > y) return;
if(x == y) {
t1[i].re(2);
t1[i][0] = 1 - p[x + 1];
t1[i][1] = p[x + 1];
ll v = x >= l ? p[x - l] : 0;
t2[i].re(2);
t2[i][0] = 1 - v;
t2[i][1] = v;
return;
}
int m = x + y >> 1;
dg(i0, x, m); dg(i1, m + 1, y);
t1[i] = t1[i0] * t1[i1];
t2[i] = t2[i0] * t2[i1];
}
V g;
V dfs(int x, int y) {
V a;
if(x > y) {
a.re(1); a[0] = 1;
return a;
}
if(x == y) {
a.re(2); a[0] = 1 - p[x]; a[1] = p[x];
return a;
}
int m = x + y >> 1;
return dfs(x, m) * dfs(m + 1, y);
}
ll a[N];
void zy(V &a, int b) {
ff(i, b, a.si) a[i - b] = a[i];
a.re(a.si - b);
}
void yy(V &a, int b) {
int sa = a.si;
a.re(sa + b);
fd(i, sa - 1, 0) a[i + b] = a[i], a[i] = 0;
}
void qz(V &a, int b) {
if(a.si > b + 1) a.re(b + 1);
}
V fz(int i, int x, int y, V f, V g, int w) {
V b; b.re(1); b[0] = 0;
if(x > y) return b;
int nw = max(0, x - (y - x));
if(nw > w) zy(f, nw - w);
qz(f, 2 * (y - x)); qz(g, (y - x));
if(x == y) {
if(x < l) {
a[x] = 0;
} else {
a[x] = (f[0] + 1) * ksm(1 - calc(x), mo - 2) % mo;
}
b.re(1); b[0] = a[x];
return b;
}
int m = x + y >> 1;
V nf = f * t2[i1], ng = g * t2[i1];
b = fz(i0, x, m, nf, ng, nw);
nf = f * t1[i0], ng = g * t1[i0];
qz(ng, y - x);
V d = b * ng; yy(d, x - nw);
nf = nf + d;
V c = fz(i1, m + 1, y, nf, ng, nw);
yy(c, m - x + 1);
return b + c;
}
int main() {
freopen("friends.in", "r", stdin);
freopen("friends.out", "w", stdout);
ntt :: build();
Init();
dg(1, 1, n);
V g;
if(l == n) g = dfs(1, 1); else {
g = dfs(2, n - l);
g.re(n + 1);
g = qni(g);
}
V f; f.re(n + 1); fo(i, 0, n) f[i] = 0;
fz(1, 1, n, f, g, 0);
pp("%lld\n", (a[n] + mo) % mo);
}
【GDOI2020模拟03.11】我的朋友们(多项式求逆+分治NTT)
标签:else turn 我的朋友 要求 alt ini 大于 题解 统计
原文地址:https://www.cnblogs.com/coldchair/p/12483517.html