Gym102870 2020-2021 “Orz Panda” Cup Programming Contest 补题记录

D. Data Structure Master and Orz Pandas
树上期望dp
列个式子跑一遍树形dp就做完了
场上剩半小时才开这题没写出来亏死

#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 7, md = 998244353;
#define ll long long
ll ksm(ll a, ll b) {a %= md;ll res = 1;while (b) {if (b & 1) {res = res * a % md;}a = a * a % md;b >>= 1;}return res % md;
}
ll inv[maxn], siz[maxn], f[maxn];
vector<int> adj[maxn];
void dfs(int u) {siz[u] = 1;for (int i = 0; i < adj[u].size(); i++) {int v = adj[u][i];dfs(v);siz[u] += siz[v];}for (int i = 0; i < adj[u].size(); i++) {int v = adj[u][i];f[u] = (f[u] + siz[v] * inv[siz[u]] % md * (f[v] + (siz[u] - siz[v] - 1) * inv[siz[u]-1] % md) % md ) % md; }
}
int rd() {int s = 0, f = 1; char c = getchar();while (c < '0' || c > '9') {if (c == '-') f = -1; c = getchar();}while (c >= '0' && c <= '9') {s = s * 10 + c - '0'; c = getchar();}return s * f;
}
int n;
int main() {n = rd();for (int i = 1; i <= n; i++) {inv[i] = ksm(i, md - 2);}for (int i = 2; i <= n; i++) {adj[rd()].push_back(i);}dfs(1);printf("%lld\n", f[1] % md);//printf("%lld\n", 4ll*inv[7] % md);return 0;
}

G. Gery’s Problem and Orz Pandas
只会O(nlogn)处理每次询问大力讨论然后加答案
然而这种树上题答案为和式形式的通常是可以化简式子然后快速统计答案的比如某经典最短路板子题
正解如下
本题要对每个询问的点对(u,v)(u,v)(u,v) 求出∑r=1ndis(u,lcar(u,v))∗dis(v,lcar(u,v))\sum_{r=1}^{n} dis(u,lca_r(u,v)) *dis(v,lca_r(u,v))r=1∑ndis(u,lcar(u,v))∗dis(v,lcar(u,v))
换根复杂度爆炸
所以我们直接在根为1的树上计算最后答案
注意到当r为lca1(u,v)lca_1(u,v)lca1(u,v)所在子树外的点时，lcar(u,v)lca_r(u,v)lcar(u,v)等于lca1(u,v)lca_1(u,v)lca1(u,v)
否则 lcar(u,v)lca_r(u,v)lcar(u,v)为u到v路径上的点
并且无论根为哪个点树上两点间的路径是唯一的
那么我们把路径分成向上的两段来统计答案
拿路径u到lca(u,v)来说
记u为v0v_0v0 则路径可表示为v0,v1,...,vqv_0,v_1,...,v_qv0,v1,...,vq
其中fa(vi−1)=vifa(v_{i-1})=v_ifa(vi−1)=vi
那么路径上的点对答案的贡献可以表示为∑i=1q(siz[vi]−siz[vi−1])×i×(q−i+dep[v]−dep[lca])\sum_{i=1}^{q}(siz[v_i]-siz[v_{i-1}]) \times i \times(q-i+dep[v]-dep[lca])i=1∑q(siz[vi]−siz[vi−1])×i×(q−i+dep[v]−dep[lca])
siz[vi]−siz[vi−1]siz[v_i]-siz[v_{i-1}]siz[vi]−siz[vi−1]表示的是以viv_ivi为根的子树内的点作为r时，满足lcar(u,v)==vilca_r(u,v)==v_ilcar(u,v)==vi的点数
化简式子我们发现iii可以表示为dep[u]−dep[vi]dep[u]-dep[v_i]dep[u]−dep[vi]
那么上式一部分改写为∑i=1q(siz[vi]−siz[vi−1])×(dep[u]−dep[vi])×(q−dep[u]+dep[vi])\sum _ {i=1}^{q} (siz[v_i]-siz[v_{i-1}])\times (dep[u]-dep[v_i]) \times(q- dep[u]+dep[v_i])i=1∑q(siz[vi]−siz[vi−1])×(dep[u]−dep[vi])×(q−dep[u]+dep[vi]) 另外一部分系数为dep[v]−dep[lca]dep[v]-dep[lca]dep[v]−dep[lca]的类似
展开以后可以把与i无关的全部提取出来
dep[u]×(q−dep[u])×∑i=1q(siz[vi]−siz[vi−1])dep[u]\times (q-dep[u])\times\sum_{i=1}^q (siz[v_i]-siz[v_{i-1}])dep[u]×(q−dep[u])×i=1∑q(siz[vi]−siz[vi−1]) +dep[u]×∑i=1q(siz[vi]−siz[vi−1])×dep[vi]+dep[u]\times\sum_{i=1}^q(siz[v_i]-siz[v_{i-1}])\times dep[v_i] +dep[u]×i=1∑q(siz[vi]−siz[vi−1])×dep[vi]
−(q−dep[u])×∑i=1q(siz[vi]−siz[vi−1])×dep[vi]-(q-dep[u])\times \sum_{i=1}^q(siz[v_i]-siz[v_{i-1}])\times dep[v_i]−(q−dep[u])×i=1∑q(siz[vi]−siz[vi−1])×dep[vi]
−∑i=1q(siz[vi]−siz[vi−1])×dep[vi]2-\sum_{i=1}^q(siz[v_i]-siz[v_{i-1}])\times dep[v_i]^2−i=1∑q(siz[vi]−siz[vi−1])×dep[vi]2

和式∑i=1q(siz[vi]−siz[vi−1])=siz[vq]−siz[v0]\sum_{i=1}^q(siz[v_i]-siz[v_{i-1}])=siz[v_q]-siz[v_0]∑i=1q(siz[vi]−siz[vi−1])=siz[vq]−siz[v0]
和式∑i=1q(siz[vi]−siz[vi−1])×dep[vi]\sum_{i=1}^q(siz[v_i]-siz[v_{i-1}])\times dep[v_i]∑i=1q(siz[vi]−siz[vi−1])×dep[vi]以及∑i=1q(siz[vi]−siz[vi−1])×dep[vi]2\sum_{i=1}^q(siz[v_i]-siz[v_{i-1}])\times dep[v_i]^2∑i=1q(siz[vi]−siz[vi−1])×dep[vi]2可以树上前缀和预处理一下
lca(u,v)lca(u,v)lca(u,v)的贡献需要特判

#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 7, md = 998244353;
#define ll long long
vector<int> adj[maxn];
ll s1[maxn], s2[maxn];
ll n, m, dfl[maxn], dep[maxn], siz[maxn], top[maxn], tot, fa[maxn], son[maxn];
int rd() {int s = 0; char c = getchar();while (c < '0' || c > '9') {c = getchar();}while (c >= '0' && c <= '9') {s = s * 10 + c - '0'; c = getchar();}return s;
}
void dfs1(int u) {siz[u] = 1;for (int i = 0; i < adj[u].size(); i++) {int v = adj[u][i];if (siz[v]) continue;dep[v] = dep[u] + 1;fa[v] = u;dfs1(v);siz[u] += siz[v];if (siz[v] > siz[son[u]]) son[u] = v;}
}
void dfs2(int u, int t) {dfl[u] = ++tot; top[u] = t;s1[u] = (s1[fa[u]] + (siz[fa[u]] - siz[u]) * dep[fa[u]] % md) % md;s2[u] = (s2[fa[u]] + dep[fa[u]] * dep[fa[u]] % md * (siz[fa[u]] - siz[u]) % md) % md;if (son[u]) {dfs2(son[u], t);}for (int i = 0; i < adj[u].size(); i++) {int v = adj[u][i];if (v != son[u] && dep[v] == dep[u]+1) {dfs2(v, v); }}
}
int getlca(int u, int v) {while (top[u] ^ top[v]) {if (dep[top[u]] < dep[top[v]]) swap(u, v);u = fa[top[u]];}if (dep[u] < dep[v]) return u;return v;
}
int getson(int u, int v) { //求u到v路径上v的儿子int res = 0; //u == v 返回0while (top[u] ^ top[v]) {if(dep[top[u]]<dep[top[v]]) swap(u,v);res = top[u]; u = fa[top[u]];}if (u == v) return res; if (dep[u] < dep[v]) return son[u];else return son[v];
}
int main() {n = rd(), m = rd();for (int i = 1; i < n; i++) {int u = rd(), v = rd();adj[u].push_back(v); adj[v].push_back(u);}dfs1(1); dfs2(1,1);for (int i = 1; i <= m; i++) {int u = rd(), v = rd();int lca = getlca(u, v);ll ans1 = 0, ans2 = 0, ans0 = 0, q = dep[u]-dep[lca];ans0 = ((dep[u]*(q-dep[u])%md*(siz[lca]-siz[u])%md+(2ll*dep[u]-q)*(s1[u]-s1[lca])%md+md-s2[u]+s2[lca])%md+md)%md;ans0 = (ans0 + dep[u]*(dep[v]-dep[lca])%md*(siz[lca]-siz[u])%md-(dep[v]-dep[lca])*(s1[u]-s1[lca])%md+md)%md;ans1 = (-siz[getson(v,lca)]*q%md*(dep[v]-dep[lca])%md+md + ans0)%md;swap(u, v); q = dep[u] - dep[lca];ans0 = ((dep[u]*(q-dep[u])%md*(siz[lca]-siz[u])%md+(2ll*dep[u]-q)*(s1[u]-s1[lca])%md+md-s2[u]+s2[lca])%md+md)%md;ans0 = (ans0 + dep[u]*(dep[v]-dep[lca])%md*(siz[lca]-siz[u])%md-(dep[v]-dep[lca])*(s1[u]-s1[lca])%md+md)%md;ans2 = (-siz[getson(v,lca)]*q%md*(dep[v]-dep[lca])%md+md + ans0)%md;ans0 = (ans1 + ans2 + (siz[1]-siz[lca])*(dep[u]-dep[lca])%md*(dep[v]-dep[lca])%md);ans0 = (ans0 - (siz[lca] - siz[getson(u,lca)] - siz[getson(v,lca)]) * (dep[u]-dep[lca]) % md * (dep[v] - dep[lca]) % md + md) % md;printf("%lld\n", ans0);}
}

E. Encryption of Orz Pandas
异或相当于每一位上进行一次模2意义的前缀和
每次前缀和相当于卷一个系数全为1的多项式
那么k次前缀和要卷的多项式的生成函数就是(1+x+x2+...+xn)k=(11−x)k(1+x+x^2+...+x^n)^k=(\frac{1}{1-x})^k(1+x+x2+...+xn)k=(1−x1)k
要求第t项的系数就泰勒展开到第t项
(ddx)t(1−x)−k=k×(k+1)×(k+2)×...×(k+t−1)(1−x)−k−t(\frac{d}{dx})^t(1-x)^{-k}=k\times(k+1)\times(k+2)\times...\times(k+t-1)(1-x)^{-k-t}(dxd)t(1−x)−k=k×(k+1)×(k+2)×...×(k+t−1)(1−x)−k−t
则(1−x)−k=∑t=0∞Ck+t−1txt(1-x)^{-k}=\sum_{t=0}^{∞}C_{k+t-1}^t x^t(1−x)−k=t=0∑∞Ck+t−1txt
第ttt项的系数就是Ck+t−1tC_{k+t-1}^{t}Ck+t−1t
在模2意义下我们判断(k+t)!,k!,t!(k+t)!,k!,t!(k+t)!,k!,t!的质因数分解中2的个数来判断最后模2结果是0还是1
然后fft算一下卷积就行了

#include<bits/stdc++.h>
using namespace std;
#define ll long long
const int maxn = 4e5+7;
const double PI = acos(-1.0);
int Bit, Lim;
ll n, k;
struct Complex{double x, y;Complex() {x = y = 0;}Complex(double _x, double _y) {x = _x, y = _y;}
}a[20][maxn], b[maxn];
Complex operator + (Complex a, Complex b) { return (Complex){a.x + b.x, a.y + b.y};}
Complex operator - (Complex a, Complex b) { return (Complex){a.x - b.x, a.y - b.y};}
Complex operator * (Complex a, Complex b) { return (Complex){a.x * b.x - a.y * b.y, a.x * b.y + a.y * b.x};}
Complex conj(Complex a) {return (Complex){a.x, -a.y};}      //共轭复数
/*
from: miskcoo's blog
假设 reverse(i) 是将二进制位反转的操作，DFT 最后一步的数组是 B，原来的数组是 A，那么 A 和 B 之间有这样的关系 B[reverse(i)]=A[i]，也就是说， B[i + 1]=A[reverse(reverse(i) + 1)]，B 中第 i 项的下一项就是将 i 反转后加 1 再反转回来 A 中的那一项，所以现在要模拟的就是从高位开始的二进制加法
反向二进制加法从最高位开始 找到第一个0 然后把这个0改成1 前面的1全部改成0
int reverse_add(int x){for (int l = 1 << bit_length; (x ^= l) < l; l >>= 1);   //如果x的当前位是0 异或后会变大 即退出return x;
}
*/
ll ans[maxn], cntk, cntt[maxn], cnttk[maxn];
ll rd(){ll s = 0, f = 1; char c = getchar();while (c > '9' || c < '0') {if (c == '-') f = -1; c = getchar();}while (c >= '0' && c <= '9') {s = s * 10 + c - '0'; c = getchar();}return s * f;
}
void fft(Complex *a, int n, int f) {   //若要进行逆变换 fft之后结果除以nfor(int i = 0, j = 0; i < n; i++) {//当前下标i 反向下标jif (i > j) swap(a[i], a[j]);    //仅交换一次 因为只有一个对应for (int l = n >> 1; (j ^= l) < l; l >>= 1);    //此时n为2的次幂}for (int i = 2; i <= n; i <<= 1) {//区间长度从小到大int m = i >> 1;   //Complex wn = (Complex) { cos(2 * PI / i), sin(2 * PI * f / i)};for (int k = 0; k < n; k += i) {//每个迭代的区间起点Complex t, w, u;w = (Complex) {1, 0};for (int j = 0; j < m; j++) {//m：半个区间t = w * a[j + m + k];u = a[k + j];a[k + j] = u + t;a[k + j + m] = u - t;w = w * wn;}}}
}
int main() {n = rd(), k = rd();for (ll i = 1; i <= n; i++) {int x = rd();for (int j = 0; j < 17; j++) {a[j][i].x = (x>>j)&1;}}Lim = 1, Bit = 0;while (Lim < n + n + 1) Lim <<= 1, Bit++;for (ll i = 2; i <= k-1ll; i <<= 1) cntk += (k-1ll)/i;for (ll i = 0; i <= n; i++) {for (ll j = 2; j <= n + k; j <<= 1ll) {cntt[i] += i / j;cnttk[i] += (i + k-1ll) / j;}}for (ll i = 0; i <= n; i++) {if (cntt[i] + cntk >= cnttk[i]) b[i].x = 1;else b[i].x = 0;}fft(b, Lim, 1);for (int j = 0; j < 17; j++) {fft(a[j], Lim, 1);for (int i = 0; i <= Lim; i++) {a[j][i] = a[j][i] * b[i];}fft(a[j], Lim, -1);for (int i = 0; i <= n; i++)  {a[j][i].x = (int)(a[j][i].x / Lim + 0.5);ans[i] += (1ll << j) * ((int)a[j][i].x%2);}}for (int i = 1; i <= n; i++) printf("%lld ", ans[i]);
}

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