ExaWizards 2019

　　AB：div 3 AB？？？

　　C：div 1 C？？？场内自闭的直接去看D。事实上是个傻逼题，注意到物品相对顺序不变，二分边界即可。

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
#define ll long long
#define N 200010
char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<'0'||c>'9')) c=getchar();return c;}
int gcd(int n,int m){return m==0?n:gcd(m,n%m);}
int read()
{
	int x=0,f=1;char c=getchar();
	while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
	while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar();
	return x*f;
}
int n,m,a[N],tot;
char s[N],b[N];
bool check(int k,int op)
{
	for (int i=1;i<=m;i++)
	if (b[i]==s[k])
	{
		if (a[i]==0) k--;
		else k++;
		if (k==0)
		{
			if (op==0) return 1;
			else return 0;
		}
		if (k==n+1)
		{
			if (op==1) return 1;
			else return 0;
		}
	}
	return 0;
}
signed main()
{
	tot=n=read(),m=read();
	scanf("%s",s+1);
	for (int i=1;i<=m;i++)
	{
		b[i]=getc();a[i]=getc()=='R';
	}
	int l=1,r=n,ans=0;
	while (l<=r)
	{
		int mid=l+r>>1;
		if (check(mid,0)) ans=mid,l=mid+1;
		else r=mid-1;
	}
	tot-=ans;
	l=1,r=n;ans=n+1;
	while (l<=r)
	{
		int mid=l+r>>1;
		if (check(mid,1)) ans=mid,r=mid-1;
		else l=mid+1;
	}
	tot-=n+1-ans;
	cout<<tot;
	return 0;
	//NOTICE LONG LONG!!!!!
}

　　D：显然对小模数取模后，大模数不会再产生影响。于是将模数从大到小排序，设f[i][j]为考虑了前i大模数后当前值是j的概率，转移考虑第i个模数是否在前缀单调栈中，若在则转移对其取模，在栈中相当于其要在比它小的所有数的前面，概率显然为1/(n-i+1)。场上莫名其妙的认为这个概率是1/(n-i+1)!，然后就自闭了。

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
#define ll long long
#define N 210
#define M 100010
#define P 1000000007
char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<'0'||c>'9')) c=getchar();return c;}
int gcd(int n,int m){return m==0?n:gcd(m,n%m);}
int read()
{
	int x=0,f=1;char c=getchar();
	while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
	while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar();
	return x*f;
}
int n,m,a[N],f[N][M],fac[N],inv[N];
signed main()
{
	freopen("d.in","r",stdin);
	freopen("d.out","w",stdout);
	n=read(),m=read();
	for (int i=1;i<=n;i++) a[i]=read();
	sort(a+1,a+n+1);reverse(a+1,a+n+1);
	fac[0]=1;for (int i=1;i<=n;i++) fac[i]=1ll*fac[i-1]*i%P;
	inv[0]=inv[1]=1;for (int i=2;i<=n;i++) inv[i]=P-1ll*(P/i)*inv[P%i]%P;
	f[0][m]=1;
	for (int i=1;i<=n;i++)
	{
		for (int j=0;j<=m;j++)
		{
			f[i][j]=(f[i][j]+1ll*f[i-1][j]*(P+1-inv[n-i+1]))%P;
			f[i][j%a[i]]=(f[i][j%a[i]]+1ll*f[i-1][j]*inv[n-i+1])%P;
		}
	}
	int ans=0;
	for (int j=0;j<=m;j++) ans=(ans+1ll*f[n][j]*fac[n]%P*(j%a[n]))%P;
	cout<<ans;
	return 0;
	//NOTICE LONG LONG!!!!!
}

　　E：最后5分钟才看这个题，然后发现是个一眼题。考虑黑白球哪个先被拿完，不妨设是白球，然后见注释。

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
#define ll long long
#define P 1000000007
#define N 200010
char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<'0'||c>'9')) c=getchar();return c;}
int gcd(int n,int m){return m==0?n:gcd(m,n%m);}
int read()
{
	int x=0,f=1;char c=getchar();
	while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
	while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar();
	return x*f;
}
int n,m,f[N],fac[N],inv[N];
int ksm(int a,int k)
{
	int s=1;
	for (;k;k>>=1,a=1ll*a*a%P) if (k&1) s=1ll*s*a%P;
	return s;
}
int Inv(int a){return ksm(a,P-2);}
int C(int n,int m){if (m>n) return 0;return 1ll*fac[n]*inv[m]%P*inv[n-m]%P;}
signed main()
{
	freopen("e.in","r",stdin);
	freopen("e.out","w",stdout);
	n=read(),m=read();
	fac[0]=fac[1]=1;for (int i=1;i<=n+m;i++) fac[i]=1ll*fac[i-1]*i%P;
	inv[0]=inv[1]=1;for (int i=2;i<=n+m;i++) inv[i]=P-1ll*(P/i)*inv[P%i]%P;
	for (int i=2;i<=n+m;i++) inv[i]=1ll*inv[i-1]*inv[i]%P;
	for (int i=m;i<n+m;i++)
	{
		int p=1ll*C(i-1,m-1)*Inv(ksm(2,i))%P;
		f[1]=(f[1]+1ll*(i-m)*Inv(i-1)%P*p)%P;
		f[i]=(f[i]+P-1ll*(i-m)*Inv(i-1)%P*p%P)%P;
		f[i+1]=(f[i+1]+p)%P;
	}
	for (int i=n;i<n+m;i++)
	{
		int p=1ll*C(i-1,n-1)*Inv(ksm(2,i))%P;
		f[1]=(f[1]+1ll*(n-1)*Inv(i-1)%P*p)%P;
		f[i]=(f[i]+P-1ll*(n-1)*Inv(i-1)%P*p%P)%P;
		f[i]=(f[i]+p)%P;
		f[i+1]=(f[i+1]+P-p)%P;
	}
	//白球是在第i次被拿完的 之前黑白球都存在 则每次拿黑白球概率均等 其概率为C(i-1,m-1)/2^i
	//考虑该情况下第j次拿黑球的概率 显然第i次不可能
	//对j<i和j>i分别考虑
	//j<i时，概率为(i-m)/(i-1)
	//j>i时，概率为1
	//若白球是最后一次被拿完的 再考虑黑球是什么时候被拿完的
	//类似
	for (int i=1;i<=n+m;i++) f[i]=(f[i]+f[i-1])%P;
	for (int i=1;i<=n+m;i++) printf("%d\n",f[i]);
	return 0;
	//NOTICE LONG LONG!!!!!
}

　　F：咕