标签：pre name pac font math 情况 notice 不可 bool
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
标签：pre name pac font math 情况 notice 不可 bool
原文地址：https://www.cnblogs.com/Gloid/p/10629779.html
