CCPC-Wannafly Winter Camp Day7
A.迷宫
有一个 n n n 个点 n − 1 n-1 n−1 条边的无向连通图迷宫，其中有些点上面有人.
现在所有人的目标都是逃离这个迷宫，而迷宫的出口是 1 号点，每一时刻，会依次发生以下的事情：
1.在点 x 上的人选择一个点 f ( x ) f(x) f(x) 作为目标，要求 f ( x ) f(x) f(x) 必须是 x x x，或者与 x x x 有边相连的点，且对于 x ≠ y x\neq y x̸=y,有 f ( x ) ≠ f ( y ) f(x)\neq f(y) f(x)̸=f(y)
2.在点 x x x 上的人移动到 f ( x ) f(x) f(x)
3.在点 1 1 1 号点上的人成功逃脱，从这个游戏里消失
现在你需要求的是：让所有人都成功逃脱至少需要多少时间.
贪心，先让深度小的节点上的人逃生。
每个人的逃生时间为 m a x ( 他 理 论 最 小 逃 生 时 间 , 上 一 个 人 的 逃 生 时 间 + 1 ) max(他理论最小逃生时间 , 上一个人的逃生时间+1) max(他理论最小逃生时间,上一个人的逃生时间+1)
#include
#define ForD(i,n) for(int i=n;i>0;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=next[p])
#define Lson (o<<1)
#define Rson ((o<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,0x3f,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define MEMx(a,b) memset(a,b,sizeof(a));
#define INF (0x3f3f3f3f)
#define F (1000000007)
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define vi vector
#define pi pair
#define PRi2D(a,n,m) For(i,n) { \
For(j,m-1) cout<
#pragma comment(linker, "/STACK:102400000,102400000")
#define ALL(x) (x).begin(),(x).end()
#define gmax(a,b) a=max(a,b);
#define gmin(a,b) a=min(a,b);
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
ll mul(ll a,ll b){return (a*b)%F;}
ll add(ll a,ll b){return (a+b)%F;}
ll sub(ll a,ll b){return ((a-b)%F+F)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}
inline int read()
{
int x=0,f=1; char ch=getchar();
while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}
while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();}
return x*f;
}
#define MAXN (112345)
int n,a[MAXN],dep[MAXN]={};
vi e[MAXN];
void dfs(int x,int fa){
dep[x]=dep[fa]+1;
for(auto v:e[x]) if(v^fa) {
dfs(v,x);
}
}
int main()
{
// freopen("A.in","r",stdin);
// freopen(".out","w",stdout);
cin>>n;For(i,n) a[i]=read();
For(i,n-1) {
int u=read(),v=read();e[u].pb(v);e[v].pb(u);
}
dfs(1,0);
vi v;
For(i,n) if(a[i]) v.pb(dep[i]-1);
sort(ALL(v));
int ans=0,p=0;
for(int x:v) {
p=max(p+1,x);
}
cout<<p<<endl;
return 0;
}
B.重新定义字典序
不会
C. 斐波那契数列
求$ sum {n=1}^R {Fib(n)&(Fib(n-1)} $
include<bits/stdc++.h>
using namespace std;
/*long long f[300];
int main()
{
f[1]=f[2]=1;
printf("%d %d\n",1,1);
printf("%d %d\n",2,1);
for(int i=3;i<=240;i++){
f[i]=f[i-2]+f[i-1];
printf("%d %lld\n",i,f[i]-(f[i]&(f[i]-1)));
}
}*/
typedef long long LL;
const int MOD=998244353;
struct Matrix{
int a[2][2];
void zero(){
for (int i = 0; i < 2; i++)
memset(a[i], 0, sizeof(int)*2);
}
void one(){
zero();
for (int i = 0; i < 2; i++)a[i][i] = 1;
}
Matrix operator * (const Matrix& x)const{
Matrix r; r.zero();
for (int i = 0; i < 2; i++){
for (int j = 0; j < 2; j++){
for (int k = 0; k < 2; k++)
r.a[i][k] = (r.a[i][k] + (LL)a[i][j] * x.a[j][k]) % MOD;
}
}
return r;
}
Matrix operator ^ (LL k)const{
Matrix r, t = *this; r.one();
for (; k; k >>= 1){
if (k & 1)r = r * t;
t = t * t;
}
return r;
}
};
int fib(LL n){
Matrix mt;
mt.a[0][0]=1;mt.a[0][1]=1;
mt.a[1][0]=1;mt.a[1][1]=0;
return (mt^n).a[1][0];
}
int main()
{
int T;
scanf("%d",&T);
while(T--){
LL n;
scanf("%lld",&n);
LL sum=fib(n+2);
sum--;
sum-=n;
sum-=n/3;
sum+=2*(n/6);
for(int i=3;i<=62;i++){
sum-=(1LL<<i)%MOD*((n+(3LL<<(i-2)))/(3LL<<(i-1)))%MOD;
}
sum=((sum%MOD)+MOD)%MOD;
printf("%lld\n",sum);
}
}
D. 二次函数
给你3个关于x的二次函数 f ( x ) , g ( x ) , h ( x ) f(x),g(x),h(x) f(x),g(x),h(x)，要求给出 x 1 , x 2 , x 3 x_1,x_2,x_3 x1,x2,x3,使得 f ( x 1 ) , g ( x 2 ) , h ( x 3 ) f(x_1),g(x_2),h(x_3) f(x1),g(x2),h(x3) 这三个数中至少有两个数相同.
系数绝对值都小于等于 1 0 4 10^4 104,
∣ x 1 ∣ , ∣ x 2 ∣ , ∣ x 3 ∣ ≤ 1 0 18 |x_1|,|x_2|,|x_3|\le 10^{18} ∣x1∣,∣x2∣,∣x3∣≤1018
分类讨论，
其实可以通过配方把一次项系数拿掉。
#includeE.线性探查法
#include
#define ForD(i,n) for(int i=n;i>0;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=next[p])
#define Lson (o<<1)
#define Rson ((o<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,0x3f,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define MEMx(a,b) memset(a,b,sizeof(a));
#define INF (0x3f3f3f3f)
#define F (1000000007)
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define vi vector
#define pi pair
#define PRi2D(a,n,m) For(i,n) { \
For(j,m-1) cout<
#pragma comment(linker, "/STACK:102400000,102400000")
#define ALL(x) (x).begin(),(x).end()
#define gmax(a,b) a=max(a,b);
#define gmin(a,b) a=min(a,b);
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
ll mul(ll a,ll b){return (a*b)%F;}
ll add(ll a,ll b){return (a+b)%F;}
ll sub(ll a,ll b){return ((a-b)%F+F)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}
inline int read()
{
int x=0,f=1; char ch=getchar();
while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}
while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();}
return x*f;
}
#define MAXN (112345+10)
int b[MAXN],a[MAXN];
set<int> S;
int get(int p) {
set<int> :: iterator it = S.lower_bound(p);
if(it==S.end()) {
it=S.lower_bound(-1);
}
return *it;
}
bool vis[MAXN]={};
int main()
{
// freopen("E.in","r",stdin);
// freopen(".out","w",stdout);
int n=read();
Rep(i,n) {
b[i]=read();
}
priority_queue< pi ,vector< pi > , greater< pi > > q;
Rep(i,n) if(b[i]%n==i) q.push(mp(b[i],i)),vis[i]=1;
int l=0;
Rep(i,n) S.insert(i);
while(!q.empty()) {
pi x=q.top();q.pop();
a[++l]=x.fi;
if(l==n) break;
S.erase(x.se);
int ad_pos=get(x.se);
if(!vis[ad_pos]) {
int c=get(b[ad_pos]%n);
if (c == ad_pos) {
q.push(mp(b[ad_pos],ad_pos));vis[ad_pos]=1;
}
}
}
PRi(a,n)
return 0;
}
F. 逆序对！
给定长度为 n 的两两不相同的整数数组 b[1…n]，定义 f(y)为：将 b 每个位置异或上 y 后，得到的新数组的逆序对个数。
需要求 ∑ i = 1 m f ( i ) \sum_{i=1}^{m}f(i) ∑i=1mf(i)
由于答案可能很大，你只需要输出答案对 998244353取模后的值
数位dp，设 d p i dp_i dpi为 1 m 1~m 1 m中第i位为1的数的个数。
排序 b b b，枚举 b i b_i bi,从高位向低位枚举x，设和 b i b_i bi的n…x+1位均相同的 b j b_j bj区间为 [ l , r ] [l,r] [l,r]。
要么这一位必须填 1 / 0 1/0 1/0，要么这一位可以填任意值（不影响答案）。
#include
#define MEM(a) memset(a,0,sizeof(a));
#define F (998244353)
typedef long long ll;
void upd(ll &a,ll b){a=(a%F+b%F)%F;}
namespace CA{
#define MAXN (32)
int a[MAXN];
ll f[MAXN][2][2][2];
//first i,jth num,k is_0qiandao,y is==n
ll dp[MAXN];
int n;
void calc(ll m) {
MEM(dp)
n=0;
while(m) a[++n]=m%2,m/=2;
reverse(a+1,a+1+n);
For(x,n) {
MEM(f)
f[0][0][0][1]=1;
Rep(i,n) {
Rep(j,2) Rep(k,2) Rep(y,2) {
if (f[i][j][k][y]) {
Rep(t,2) {
int nk=k|(t>0);
int ny=(y&&t==a[i+1]);
if (y&&t>a[i+1]) continue;
if(x==i+1 && !t) continue;
upd(f[i+1][t][nk][ny],f[i][j][k][y]);
}
}
}
}
int k=1;
Rep(j,2) Rep(y,2) {
upd(dp[x],f[n][j][k][y]);
}
}
reverse(dp+1,dp+1+n);
For(i,n) dp[i-1]=dp[i];
dp[n]=0;
}
}
/*int main()
{
// freopen("E.in","r",stdin);
// freopen(".out","w",stdout);
CA::calc(4);
return 0;
}*/
#include G.抢红包机器人
#include
#define ForD(i,n) for(int i=n;i>0;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=next[p])
#define Lson (o<<1)
#define Rson ((o<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,0x3f,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define MEMx(a,b) memset(a,b,sizeof(a));
#define INF (0x3f3f3f3f)
#define F (1000000007)
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define vi vector
#define pi pair
#define PRi2D(a,n,m) For(i,n) { \
For(j,m-1) cout<
#pragma comment(linker, "/STACK:102400000,102400000")
#define ALL(x) (x).begin(),(x).end()
#define gmax(a,b) a=max(a,b);
#define gmin(a,b) a=min(a,b);
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
ll mul(ll a,ll b){return (a*b)%F;}
ll add(ll a,ll b){return (a+b)%F;}
ll sub(ll a,ll b){return ((a-b)%F+F)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}
inline int read()
{
int x=0,f=1; char ch=getchar();
while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}
while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();}
return x*f;
}
#define MAXN (101)
queue<int> q;
bool b[MAXN]={};
int g[MAXN][MAXN]={},f[MAXN][MAXN]={};
int main()
{
// freopen("g.in","r",stdin);
// freopen(".out","w",stdout);
int n=read(),m=read();
For(i,m) {
int k=read();
For(j,k) {
f[i][j]=read();
}
For(j,k-1){
int l=j+1;
g[f[i][l]][f[i][j]]=1;
}
}
int Ans=1e9;
For(xnode,n) {
MEM(b)
b[xnode]=1;q.push(xnode);
while(!q.empty()) {
int x=q.front();
q.pop();
For(i,n) if(!b[i] && g[x][i]) b[i]=1,q.push(i);
}
int ans=0;
For(i,n) ans+= b[i];
// cout<
gmin(Ans,ans)
}
cout<<Ans<<endl;
return 0;
}
H.同构
给定 n，我们定义一张无向图是好的，当且仅当它无重边无自环，且每个点的度数都是 n-3。你需要求出最多能找出多少张好的图，使得它们两两不同构.
输出答案对 998244353 取模后的值
n ≤ 1 e 5 n\le 1e5 n≤1e5
补图，转成环，划分数
#includeI. 集合
J.强壮的排列
我们定义一个 1... n 1...n 1...n 的排列 p 是强壮的，当且仅当 n 是奇数，且对于所有 1 ≤ i ≤ n − 1 2 1\leq i\leq \frac{n-1}{2} 1≤i≤2n−1有 p [ 2 i ] > m a x ( p [ 2 i − 1 ] , p [ 2 i + 1 ] ) p[2i]>max(p[2i−1],p[2i+1]) p[2i]>max(p[2i−1],p[2i+1])
现在你需要计算，有几个长度为 n 的排列是强壮的，由于答案可能较大，你只需要输出答案对 998244353取模后的值
经过各种计算发现答案为 t a n ( x ) tan(x) tan(x)的生成函数第n项系数
可以用伯努利数计算这个值
#include
#define ForD(i,n) for(int i=n;i>0;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=next[p])
#define Lson (o<<1)
#define Rson ((o<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,0x3f,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define MEMx(a,b) memset(a,b,sizeof(a));
#define INF (0x3f3f3f3f)
#define F (998244353)
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define vi vector
#define pi pair
#define PRi2D(a,n,m) For(i,n) { \
For(j,m-1) cout<
#pragma comment(linker, "/STACK:102400000,102400000")
#define ALL(x) (x).begin(),(x).end()
#define gmax(a,b) a=max(a,b);
#define gmin(a,b) a=min(a,b);
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
ll mul(ll a,ll b){return (a*b)%F;}
ll add(ll a,ll b){return (a+b)%F;}
ll sub(ll a,ll b){return ((a-b)%F+F)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}
inline int read()
{
int x=0,f=1; char ch=getchar();
while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}
while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();}
return x*f;
}
#define MAXN (212345)
ll inj[MAXN],jie[MAXN];
void pre(int n) {
jie[0]=1;For(i,n) jie[i]=mul(jie[i-1],i);
inj[0]=inj[1]=1;Fork(i,2,n) inj[i]=(F-(F/i))*inj[F%i]%F;
For(i,n) inj[i]=mul(inj[i],inj[i-1]);
}
ll p=F;
inline int pow2(int a,ll b) //a^b mod p
{
if (b==0) return 1%p;
int c=pow2(a,b/2);
c=(ll)c*c%p;
if (b&1) c=(ll)c*a%p;
return c;
}
ll inv(ll a) { //gcd(a,p)=1
return pow2(a,p-2);
}
#undef MAXN
namespace POLY{
#define MAXN 262144
const int MOD = 998244353, ROOT = 3;
typedef vector<int> Poly;
inline int add(int a, int b){ return (a + b) % MOD; }
inline int sub(int a, int b){ return (a - b + MOD) % MOD; }
inline int mul(int a, int b){ return (long long)a * b % MOD; }
int power(int a, int b){
int ret = 1;
for (int t = a; b; b >>= 1){
if (b & 1)ret = mul(ret, t);
t = mul(t, t);
}
return ret;
}
int pow2(int n){
int ret = 1;
while (ret < n)ret *= 2;
return ret;
}
void clean(Poly& p){
while (!p.empty() && !p.back())p.pop_back();
}
Poly cut(const Poly& p, int n){
return Poly(p.begin(), p.begin() + n);
}
Poly expand(const Poly& p, int n){
Poly ret(n);
copy(p.begin(), p.end(), ret.begin());
return ret;
}
Poly operator + (Poly p, const Poly& p2){
p.resize(max(p.size(), p2.size()));
for (int i = 0; i < (int)p2.size(); i++)
p[i] = add(p[i], p2[i]);
return p;
}
Poly operator - (Poly p, const Poly& p2){
p.resize(max(p.size(), p2.size()));
for (int i = 0; i < (int)p2.size(); i++)
p[i] = sub(p[i], p2[i]);
return p;
}
Poly operator * (Poly p, int x)
{
for (int i = 0; i < (int)p.size(); i++)
p[i] = mul(p[i], x);
return p;
}
void dot(Poly& p, const Poly& p2){
for (int i = 0; i < (int)p.size(); i++)
p[i] = mul(p[i], p2[i]);
}
void fft(Poly& a, bool rev){
static int g[MAXN + 1], w[MAXN] ;
if (!g[0]){
w[0] = 1;
g[0] = 1; g[1] = power(ROOT, (MOD - 1) / MAXN);
for (int i = 2; i < MAXN; i++)
g[i] = mul(g[i - 1], g[1]);
}
int len = a.size(), d = MAXN / len;
for (int i = 1, j = len / 2; i < len; i++){
w[i] = g[d * (rev ? len - i : i)];
if (i < j) swap(a[i], a[j]);
for (int k = len; j < k; k >>= 1, j ^= k);
}
for (int s = 1; s < len; s <<= 1){
int step = len / s / 2;
for (int j = 0; j < len; j += 2 * s){
for (int k = j, wk = 0; k < j + s; k++, wk += step){
int t = mul(w[wk], a[k + s]);
a[k + s] = sub(a[k], t); a[k] = add(a[k], t);
}
}
}
if (rev){
int t = MOD / len; t = mul(t, t * len);
for (int i = 0; i < len; i++)
a[i] = mul(a[i], t);
}
}
Poly operator * (const Poly& p1, const Poly& p2)
{
if (p1.empty() || p2.empty())return Poly();
int len = pow2(p1.size() + p2.size());
Poly p3 = expand(p1, len), p4 = expand(p2, len);
fft(p3, 0); fft(p4, 0);
dot(p3, p4); fft(p3, 1);
p3.resize(p1.size() + p2.size() - 1);
return p3;
}
Poly inv(const Poly& p)
{
int len = 2 * pow2(p.size());
Poly a, b(len), exa = expand(p, len);
b[0] = power(p[0], MOD - 2);
for (int i = 4; i <= len; i *= 2){
a = cut(exa, i / 2);
a.resize(i); b.resize(i);
fft(a, 0); fft(b, 0); dot(a, b);
dot(b, Poly(i, 2) - a);
fft(b, 1); b.resize(i / 2);
}
b.resize(p.size());
return b;
}
int fact[MAXN], factInv[MAXN];
void init()
{
fact[0] = 1;
for (int i = 1; i < MAXN; i++)
fact[i] = mul(fact[i - 1], i);
factInv[MAXN - 1] = power(fact[MAXN - 1], MOD - 2);
for (int i = MAXN - 1; i; i--)
factInv[i - 1] = mul(factInv[i], i);
}
Poly bernoulli(int n)
{
Poly p(n + 1);
for (int i = 0; i <= n; i++)
p[i] = factInv[i + 1];
return inv(p);
}
}
//a(n) = abs[c(2*n-1)] where c(n)= 2^(n+1) * (1-2^(n+1)) * Ber(n+1)/(n+1)
int main()
{
// freopen("B.in","r",stdin);
// freopen(".out","w",stdout);
pre(2e5+10);
POLY::init();
vi v=POLY::bernoulli(100010);
int T=read();
Rep(i,100010) v[i]=v[i]*(ll)jie[i]%F;
while(T--) {
ll n=read();n=(n+1)/2;n=n*2-1;
ll t=pow2(2,n+1),t2=sub(1,pow2(2,n+1));
ll t3=inv(n+1);
ll t4=v[n+1];
ll t5=t*t2%F*t3%F*t4%F;
if((n+1)%4==0 ||n+1==1) t5=sub(F,t5);
if(n+1>=1) t5=sub(F,t5);
cout<<t5%F<<endl;
}
return 0;
}