Div 2的比赛，前四题还有那么多人过，应该是SB题，就不讲了。
这场比赛一堆计数题，很舒服。（虽然我没打）
E. The Top Scorer
其实这题也不难，不知道为什么这么少人过。
考虑枚举那人的分数和有多少人和他同分，推一下就会发现我们只需要知道\(calc(sum,n,top)\)表示\(sum\)分，分给\(n\)个人，分数小于\(top\)，的方案数。
好像不是很好直接搞，考虑容斥，枚举一下至少有几个人不满足条件即可。
#include<bits/stdc++.h>
namespace my_std{
using namespace std;
#define pii pair<int,int>
#define fir first
#define sec second
#define MP make_pair
#define rep(i,x,y) for (int i=(x);i<=(y);i++)
#define drep(i,x,y) for (int i=(x);i>=(y);i--)
#define go(x) for (int i=head[x];i;i=edge[i].nxt)
#define mod (ll(998244353))
#define sz 10101
typedef long long ll;
typedef double db;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
template<typename T>inline T rnd(T l,T r) {return uniform_int_distribution<T>(l,r)(rng);}
template<typename T>inline void read(T& t)
{
t=0;char f=0,ch=getchar();double d=0.1;
while(ch>'9'||ch<'0') f|=(ch=='-'),ch=getchar();
while(ch<='9'&&ch>='0') t=t*10+ch-48,ch=getchar();
if(ch=='.'){ch=getchar();while(ch<='9'&&ch>='0') t+=d*(ch^48),d*=0.1,ch=getchar();}
t=(f?-t:t);
}
template<typename T,typename... Args>inline void read(T& t,Args&... args){read(t); read(args...);}
void file()
{
#ifndef ONLINE_JUDGE
freopen("a.txt","r",stdin);
#endif
}
#ifdef mod
ll ksm(ll x,int y){ll ret=1;for (;y;y>>=1,x=x*x%mod) if (y&1) ret=ret*x%mod;return ret;}
ll inv(ll x){return ksm(x,mod-2);}
#else
ll ksm(ll x,int y){ll ret=1;for (;y;y>>=1,x=x*x) if (y&1) ret=ret*x;return ret;}
#endif
// inline ll mul(ll a,ll b){ll d=(ll)(a*(double)b/mod+0.5);ll ret=a*b-d*mod;if (ret<0) ret+=mod;return ret;}
}
using namespace my_std;
ll fac[sz],_fac[sz];
void init(){fac[0]=_fac[0]=1;rep(i,1,sz-1) _fac[i]=inv(fac[i]=fac[i-1]*i%mod);}
ll C(int n,int m){return n>=m&&m>=0?fac[n]*_fac[m]%mod*_fac[n-m]%mod:0;}
int n,r,s;
ll calc(int sum,int n,int top) // sum points for n people , < top
{
if (!n) return sum==0;
ll ret=0;
rep(i,0,n)
{
if (i*top>sum) return ret;
int cur=sum-i*top;
ret=(ret+1ll*((i&1)?-1:1)*C(cur+n-1,n-1)*C(n,i)%mod+mod)%mod;
}
return ret;
}
int main()
{
file();
init();
read(n,s,r);
ll tot=C(s-r+n-1,n-1);
ll ans=0;
rep(i,r,s)
{
rep(j,0,n-1)
{
int rest=s-i-i*j;if (rest<0) break;
ans=(ans+calc(rest,n-j-1,i)*C(n-1,j)%mod*inv(j+1)%mod)%mod;
}
}
cout<<ans*inv(tot)%mod;
return 0;
}
F. Inversion Expectation
很容易想到把各个部分的贡献拆开来算。
分成三个部分：已知对已知、未知对未知、未知对已知。
前两个都很好搞，第三个考虑期望的线性性（虽然我不知道那是啥），随便搞搞就好了。
#include<bits/stdc++.h>
clock_t t=clock();
namespace my_std{
using namespace std;
#define pii pair<int,int>
#define fir first
#define sec second
#define MP make_pair
#define rep(i,x,y) for (int i=(x);i<=(y);i++)
#define drep(i,x,y) for (int i=(x);i>=(y);i--)
#define go(x) for (int i=head[x];i;i=edge[i].nxt)
#define templ template<typename T>
#define mod 998244353ll
#define sz 200220
typedef long long ll;
typedef double db;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
templ inline T rnd(T l,T r) {return uniform_int_distribution<T>(l,r)(rng);}
templ inline bool chkmax(T &x,T y){return x<y?x=y,1:0;}
templ inline bool chkmin(T &x,T y){return x>y?x=y,1:0;}
templ inline void read(T& t)
{
t=0;char f=0,ch=getchar();double d=0.1;
while(ch>'9'||ch<'0') f|=(ch=='-'),ch=getchar();
while(ch<='9'&&ch>='0') t=t*10+ch-48,ch=getchar();
if(ch=='.'){ch=getchar();while(ch<='9'&&ch>='0') t+=d*(ch^48),d*=0.1,ch=getchar();}
t=(f?-t:t);
}
template<typename T,typename... Args>inline void read(T& t,Args&... args){read(t); read(args...);}
void file()
{
#ifndef ONLINE_JUDGE
freopen("a.in","r",stdin);
#endif
}
inline void chktime()
{
#ifndef ONLINE_JUDGE
cout<<(clock()-t)/1000.0<<'\n';
#endif
}
#ifdef mod
ll ksm(ll x,int y){ll ret=1;for (;y;y>>=1,x=x*x%mod) if (y&1) ret=ret*x%mod;return ret;}
ll inv(ll x){return ksm(x,mod-2);}
#else
ll ksm(ll x,int y){ll ret=1;for (;y;y>>=1,x=x*x) if (y&1) ret=ret*x;return ret;}
#endif
// inline ll mul(ll a,ll b){ll d=(ll)(a*(double)b/mod+0.5);ll ret=a*b-d*mod;if (ret<0) ret+=mod;return ret;}
}
using namespace my_std;
int n,m;
int a[sz];
bool vis[sz];
ll tr[sz];
void add(int x,int y){while (x<=n) (tr[x]+=y)%=mod,x+=(x&(-x));}
int query(int x){ll ret=0;while (x) (ret+=tr[x])%=mod,x-=(x&(-x));return ret;}
ll fac[sz];
int main()
{
file();
cin>>n;
rep(i,1,n)
{
cin>>a[i];
if (a[i]==-1) ++m;
}
ll ans=0;
rep(i,1,n) if (a[i]!=-1) (ans+=query(n)-query(a[i]))%=mod,add(a[i],1);
fac[0]=1;rep(i,1,n) fac[i]=fac[i-1]*i%mod;
(ans*=fac[m])%=mod;
(ans+=fac[m]*inv(4)%mod*(1ll*m*(m-1)%mod)%mod)%=mod;
int cnt=0;
rep(i,1,n)
if (a[i]!=-1) (ans+=fac[m-1]*(1ll*cnt*(m+query(a[i])-a[i])%mod+1ll*(m-cnt)*(a[i]-query(a[i]))%mod)%mod)%=mod;
else ++cnt;
cout<<ans*inv(fac[cnt])%mod;
return 0;
}
G. Lucky Tickets
很容易想到枚举两边有多少分。
考虑一个DP：\(dp_{i,j}\)表示前\(i\)位的和为\(j\)的方案数，转移方程显然。
感受一下，这东西就是一个多项式快速幂，就做完了。
#include<bits/stdc++.h>
clock_t t=clock();
namespace my_std{
using namespace std;
#define pii pair<int,int>
#define fir first
#define sec second
#define MP make_pair
#define rep(i,x,y) for (int i=(x);i<=(y);i++)
#define drep(i,x,y) for (int i=(x);i>=(y);i--)
#define go(x) for (int i=head[x];i;i=edge[i].nxt)
#define templ template<typename T>
#define sz 8010100
#define mod 998244353ll
typedef long long ll;
typedef double db;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
templ inline T rnd(T l,T r) {return uniform_int_distribution<T>(l,r)(rng);}
templ inline bool chkmax(T &x,T y){return x<y?x=y,1:0;}
templ inline bool chkmin(T &x,T y){return x>y?x=y,1:0;}
templ inline void read(T& t)
{
t=0;char f=0,ch=getchar();double d=0.1;
while(ch>'9'||ch<'0') f|=(ch=='-'),ch=getchar();
while(ch<='9'&&ch>='0') t=t*10+ch-48,ch=getchar();
if(ch=='.'){ch=getchar();while(ch<='9'&&ch>='0') t+=d*(ch^48),d*=0.1,ch=getchar();}
t=(f?-t:t);
}
template<typename T,typename... Args>inline void read(T& t,Args&... args){read(t); read(args...);}
void file()
{
#ifndef ONLINE_JUDGE
freopen("a.in","r",stdin);
#endif
}
inline void chktime()
{
#ifndef ONLINE_JUDGE
cout<<(clock()-t)/1000.0<<'\n';
#endif
}
#ifdef mod
ll ksm(ll x,int y){ll ret=1;for (;y;y>>=1,x=x*x%mod) if (y&1) ret=ret*x%mod;return ret;}
ll inv(ll x){return ksm(x,mod-2);}
#else
ll ksm(ll x,int y){ll ret=1;for (;y;y>>=1,x=x*x) if (y&1) ret=ret*x;return ret;}
#endif
// inline ll mul(ll a,ll b){ll d=(ll)(a*(double)b/mod+0.5);ll ret=a*b-d*mod;if (ret<0) ret+=mod;return ret;}
}
using namespace my_std;
int limit,r[sz];
void NTT_init(int n)
{
limit=1;int l=-1;
while (limit<=n+n) limit<<=1,++l;
rep(i,0,limit-1) r[i]=(r[i>>1]>>1)|((i&1)<<l);
}
void NTT(ll *a,int type)
{
rep(i,0,limit-1) if (i<r[i]) swap(a[i],a[r[i]]);
rep(i,0,limit-1) a[i]%=mod;
for (int mid=1;mid<limit;mid<<=1)
{
ll Wn=ksm(3,(mod-1)/mid>>1);if (type==-1) Wn=inv(Wn);
for (int j=0,len=mid<<1;j<limit;j+=len)
{
ll w=1;
for (int k=0;k<mid;k++,w=w*Wn%mod)
{
ll x=a[j+k],y=a[j+k+mid]*w;
a[j+k]=(x+y)%mod;a[j+k+mid]=(1ll*mod*mod-y+x)%mod;
}
}
}
if (type==1) return;
ll I=inv(limit);
rep(i,0,limit-1) a[i]=a[i]*I%mod;
}
int n,K;
ll a[sz];
int main()
{
file();
read(n,K);
int x;
rep(i,1,K) read(x),a[x]=1;
NTT_init(n*10);
NTT(a,1);
rep(i,0,limit) a[i]=ksm(a[i],n/2);
NTT(a,-1);
ll ans=0;
rep(i,0,n*10) (ans+=a[i]*a[i]%mod)%=mod;
cout<<ans;
return 0;
}