G. The Symbolic Tree
一棵拥有 n 个顶点的有根树。树根为顶点 1，
根节点上写的整数是 K。
对于除根节点以外的每个顶点，该顶点上写的整数不大于其父节点上写的正整数。
求在树的顶点上写整数的所有可能方案数对 998244353 取模后的余数。
当且仅当存在某个顶点，其上写的整数在两种方案中不同时，这两种写整数的方案被视为不同。
n≤3000,k≤1e9n\le 3000 ,k\le 1e9n≤3000,k≤1e9
先计算出k≤nk\le nk≤n 的答案，然后用拉格朗日推导后面的
cpp
#include<bits/stdc++.h>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define ForkD(i,k,n) for(int i=n;i>=k;i--)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;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<int>
#define pi pair<int,int>
#define SI(a) ((a).size())
#define Pr(kcase,ans) printf("Case #%d: %lld\n",kcase,ans);
#define PRi(a,n) For(i,n-1) cout<<a[i]<<' '; cout<<a[n]<<endl;
#define PRi2D(a,n,m) For(i,n) { \
For(j,m-1) cout<<a[i][j]<<' ';\
cout<<a[i][m]<<endl; \
}
#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);
#define sqr(x) (x*x)
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;}
ll gcd(ll a,ll b){return (!a)?b:a%b;}
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;
}
template<int MOD, int RT> struct mint {
static const int mod = MOD;
static constexpr mint rt() { return RT; } // primitive root for FFT
int v; explicit operator int() const { return v; } // explicit -> don't silently convert to int
mint():v(0) {}
mint(ll _v) { v = int((-MOD < _v && _v < MOD) ? _v : _v % MOD);
if (v < 0) v += MOD; }
bool operator==(const mint& o) const {
return v == o.v; }
friend bool operator!=(const mint& a, const mint& b) {
return !(a == b); }
friend bool operator<(const mint& a, const mint& b) {
return a.v < b.v; }
mint& operator+=(const mint& o) {
if ((v += o.v) >= MOD) v -= MOD;
return *this; }
mint& operator-=(const mint& o) {
if ((v -= o.v) < 0) v += MOD;
return *this; }
mint& operator*=(const mint& o) {
v = int((ll)v*o.v%MOD); return *this; }
mint& operator/=(const mint& o) { return (*this) *= inv(o); }
friend mint pow(mint a, ll p) {
mint ans = 1; assert(p >= 0);
for (; p; p /= 2, a *= a) if (p&1) ans *= a;
return ans; }
friend mint inv(const mint& a) { assert(a.v != 0);
return pow(a,MOD-2); }
// mint inv() { assert(v != 0);
// return pow(v,MOD-2); }
mint operator-() const { return mint(-v); }
mint& operator++() { return *this += 1; }
mint& operator--() { return *this -= 1; }
friend mint operator+(mint a, const mint& b) { return a += b; }
friend mint operator-(mint a, const mint& b) { return a -= b; }
friend mint operator*(mint a, const mint& b) { return a *= b; }
friend mint operator/(mint a, const mint& b) { return a /= b; }
};
const int MOD=998244353;
using mi = mint<MOD,5>; // 5 is primitive root for both common mods
int n;
#define MAXN (3010)
mi f[MAXN][MAXN]={},s[MAXN][MAXN]={};
vector<int> v[MAXN];
void dfs(int x,int fa) {
bool is_leaf=true;
for (auto u:v[x])
if (u!=fa) {
is_leaf=false;
dfs(u,x);
}
if (is_leaf) {
For(i,n) f[x][i]=1,s[x][i]=i;
}else{
For(i,n){
mi pord=1;
for(auto u:v[x]){
if (u==fa) continue;
pord*=s[u][i];
}
f[x][i]=pord;
s[x][i]=s[x][i-1]+f[x][i];
}
}
}
mi lagrange_prediction(vector<mi> y, long long k) {
int n = y.size() - 1;
if (k <= n) return y[k];
mi ans = 0;
for (int i = 0; i <= n; i++) {
mi num = 1, den = 1;
for (int j = 0; j <= n; j++) {
if (j == i) continue;
num = num * (k - j);
den = den * (i - j);
}
ans = ans + y[i] * num / den;
}
return ans;
}
int main()
{
// freopen(".in","r",stdin);
// freopen(".out","w",stdout);
ios::sync_with_stdio(false);
cin.tie(0);
cin>>n;ll m;cin>>m;
For(i,n-1) {
int x,y;
cin>>x>>y;
v[x].pb(y); v[y].pb(x);
}
dfs(1,-1);
if(m+1<=n)
cout<<f[1][m+1].v<<endl;
else {
vector<mi> vec(n);
For(i,n) vec[i-1]=f[1][i];
cout<<lagrange_prediction(vec,m).v<<endl;
}
}L. Make Many KUPC
给定一个长度为 n 的字符串 S，由大写英文字母组成。你可以对
执行任意次数以下操作：
选择一个四元组整数 (i,j,k,l)
，满足 SiSjSkSl=="KUPC"S_iS_jS_kS_l=="KUPC"SiSjSkSl=="KUPC"
将它们 全部替换为X，并获得i∗j∗k∗li*j*k*li∗j∗k∗l 日元的收益。
求总共能获得的最大收益，结果对 998244353 取模。
cpp
#include<bits/stdc++.h>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define ForkD(i,k,n) for(int i=n;i>=k;i--)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;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<int>
#define pi pair<int,int>
#define SI(a) ((a).size())
#define Pr(kcase,ans) printf("Case #%d: %lld\n",kcase,ans);
#define PRi(a,n) For(i,n-1) cout<<a[i]<<' '; cout<<a[n]<<endl;
#define PRi2D(a,n,m) For(i,n) { \
For(j,m-1) cout<<a[i][j]<<' ';\
cout<<a[i][m]<<endl; \
}
#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);
#define sqr(x) (x*x)
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;}
ll gcd(ll a,ll b){return (!a)?b:a%b;}
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;
}
template<int MOD, int RT> struct mint {
static const int mod = MOD;
static constexpr mint rt() { return RT; } // primitive root for FFT
int v; explicit operator int() const { return v; } // explicit -> don't silently convert to int
mint():v(0) {}
mint(ll _v) { v = int((-MOD < _v && _v < MOD) ? _v : _v % MOD);
if (v < 0) v += MOD; }
bool operator==(const mint& o) const {
return v == o.v; }
friend bool operator!=(const mint& a, const mint& b) {
return !(a == b); }
friend bool operator<(const mint& a, const mint& b) {
return a.v < b.v; }
mint& operator+=(const mint& o) {
if ((v += o.v) >= MOD) v -= MOD;
return *this; }
mint& operator-=(const mint& o) {
if ((v -= o.v) < 0) v += MOD;
return *this; }
mint& operator*=(const mint& o) {
v = int((ll)v*o.v%MOD); return *this; }
mint& operator/=(const mint& o) { return (*this) *= inv(o); }
friend mint pow(mint a, ll p) {
mint ans = 1; assert(p >= 0);
for (; p; p /= 2, a *= a) if (p&1) ans *= a;
return ans; }
friend mint inv(const mint& a) { assert(a.v != 0);
return pow(a,MOD-2); }
mint operator-() const { return mint(-v); }
mint& operator++() { return *this += 1; }
mint& operator--() { return *this -= 1; }
friend mint operator+(mint a, const mint& b) { return a += b; }
friend mint operator-(mint a, const mint& b) { return a -= b; }
friend mint operator*(mint a, const mint& b) { return a *= b; }
friend mint operator/(mint a, const mint& b) { return a /= b; }
};
const int MOD=998244353;
using mi = mint<MOD,5>; // 5 is primitive root for both common mods
namespace simp {
vector<mi> fac,ifac,invn;
void check(int x) {
if (fac.empty()) {
fac={mi(1),mi(1)};
ifac={mi(1),mi(1)};
invn={mi(0),mi(1)};
}
while (SI(fac)<=x) {
int n=SI(fac),m=SI(fac)*2;
fac.resize(m);
ifac.resize(m);
invn.resize(m);
for (int i=n;i<m;i++) {
fac[i]=fac[i-1]*mi(i);
invn[i]=mi(MOD-MOD/i)*invn[MOD%i];
ifac[i]=ifac[i-1]*invn[i];
}
}
}
mi gfac(int x) {
assert(x>=0);
check(x); return fac[x];
}
mi ginv(int x) {
assert(x>0);
check(x); return invn[x];
}
mi gifac(int x) {
assert(x>=0);
check(x); return ifac[x];
}
mi binom(int n,int m) {
if (m < 0 || m > n) return mi(0);
return gfac(n)*gifac(m)*gifac(n - m);
}
}
int main()
{
// freopen(".in","r",stdin);
// freopen(".out","w",stdout);
ios::sync_with_stdio(false);
cin.tie(0);
int n;
cin>>n;
string s;
cin>>s;
set<int> S[26];
Rep(i,n) {
S[s[i]-'A'].insert(i+1);
}
ll ans=0;
while(1) {
auto it4=S['C'-'A'].end();
if(it4==S['C'-'A'].begin()) break;
it4--;
ll t4=*it4;
auto it3=S['P'-'A'].upper_bound(t4);
if(it3==S['P'-'A'].begin()) break;
it3--;
ll t3=*it3;
auto it2=S['U'-'A'].upper_bound(t3);
if(it2==S['U'-'A'].begin()) break;
it2--;
ll t2=*it2;
auto it1=S['K'-'A'].upper_bound(t2);
if(it1==S['K'-'A'].begin()) break;
it1--;
ll t1=*it1;
ans+=(t1)*(t2)%F*(t3)%F*(t4);
ans%=998244353;
S['K'-'A'].erase(t1);
S['U'-'A'].erase(t2);
S['P'-'A'].erase(t3);
S['C'-'A'].erase(t4);
}
cout<<ans;
return 0;
}O Xor Triangle
求满足 1≤a,b<2n1 \le a,b \lt 2^n1≤a,b<2n 且满足以下条件的整数对 (a,b)(a,b)(a,b)
的数量，结果对质数 998244353 取模。
存在一个以 a,b,a xor b 为边长的非退化三角形。
oeis
cpp
#include<bits/stdc++.h>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define ForkD(i,k,n) for(int i=n;i>=k;i--)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;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<int>
#define pi pair<int,int>
#define SI(a) ((a).size())
#define Pr(kcase,ans) printf("Case #%d: %lld\n",kcase,ans);
#define PRi(a,n) For(i,n-1) cout<<a[i]<<' '; cout<<a[n]<<endl;
#define PRi2D(a,n,m) For(i,n) { \
For(j,m-1) cout<<a[i][j]<<' ';\
cout<<a[i][m]<<endl; \
}
#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);
#define sqr(x) (x*x)
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;}
ll gcd(ll a,ll b){return (!a)?b:a%b;}
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;
}
//4**(n-1) - 3*3**(n-1) + 3*2**(n-1) - 1)
template<int MOD, int RT> struct mint {
static const int mod = MOD;
static constexpr mint rt() { return RT; } // primitive root for FFT
int v; explicit operator int() const { return v; } // explicit -> don't silently convert to int
mint():v(0) {}
mint(ll _v) { v = int((-MOD < _v && _v < MOD) ? _v : _v % MOD);
if (v < 0) v += MOD; }
bool operator==(const mint& o) const {
return v == o.v; }
friend bool operator!=(const mint& a, const mint& b) {
return !(a == b); }
friend bool operator<(const mint& a, const mint& b) {
return a.v < b.v; }
mint& operator+=(const mint& o) {
if ((v += o.v) >= MOD) v -= MOD;
return *this; }
mint& operator-=(const mint& o) {
if ((v -= o.v) < 0) v += MOD;
return *this; }
mint& operator*=(const mint& o) {
v = int((ll)v*o.v%MOD); return *this; }
mint& operator/=(const mint& o) { return (*this) *= inv(o); }
friend mint pow(mint a, ll p) {
mint ans = 1; assert(p >= 0);
for (; p; p /= 2, a *= a) if (p&1) ans *= a;
return ans; }
friend mint inv(const mint& a) { assert(a.v != 0);
return pow(a,MOD-2); }
mint operator-() const { return mint(-v); }
mint& operator++() { return *this += 1; }
mint& operator--() { return *this -= 1; }
friend mint operator+(mint a, const mint& b) { return a += b; }
friend mint operator-(mint a, const mint& b) { return a -= b; }
friend mint operator*(mint a, const mint& b) { return a *= b; }
friend mint operator/(mint a, const mint& b) { return a /= b; }
};
const int MOD=998244353;
using mi = mint<MOD,5>; // 5 is primitive root for both common mods
namespace simp {
vector<mi> fac,ifac,invn;
void check(int x) {
if (fac.empty()) {
fac={mi(1),mi(1)};
ifac={mi(1),mi(1)};
invn={mi(0),mi(1)};
}
while (SI(fac)<=x) {
int n=SI(fac),m=SI(fac)*2;
fac.resize(m);
ifac.resize(m);
invn.resize(m);
for (int i=n;i<m;i++) {
fac[i]=fac[i-1]*mi(i);
invn[i]=mi(MOD-MOD/i)*invn[MOD%i];
ifac[i]=ifac[i-1]*invn[i];
}
}
}
mi gfac(int x) {
assert(x>=0);
check(x); return fac[x];
}
mi ginv(int x) {
assert(x>0);
check(x); return invn[x];
}
mi gifac(int x) {
assert(x>=0);
check(x); return ifac[x];
}
mi binom(int n,int m) {
if (m < 0 || m > n) return mi(0);
return gfac(n)*gifac(m)*gifac(n - m);
}
}
int main()
{
// freopen(".in","r",stdin);
// freopen(".out","w",stdout);
ios::sync_with_stdio(false);
cin.tie(0);
// int n=read();
// int S=1<<n;
// int t=0;
// For(i,S-1) For(j,S-1) {
// int k=i^j;
// if(i+j>k && i+k>j && j+k>i) {
// ++t;
// }
// }
// cout<<t;
ll n;cin>>n;++n;
cout<<(pow((mi)4,(n-1)) - (mi)3*pow((mi)3,n-1) + 3*pow((mi)2,n-1) - (mi)1).v;
return 0;
}