Codeforces Educational Codeforces Round 57 题解

Posted 2021-02-10 p-b-p-b

传送门

---

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<<‘
‘;
        #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<<‘
‘;
        #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;
}