T1 Scheduling
If you consider all numbers are different, then it must only one feasible, you need to \ (n! \) Times
If several identical, to consider specific magnitude relationship, in addition to the last number of the same number are arranged within each of
Consider how to add a new number, the more the average clearly better simulate what you can
Complexity \ (O (Tn \ log n ) \)
#include <bits/stdc++.h>
#define N 200005
#define M 10000005
#define mod 998244353
#define getchar nc
using namespace std;
inline char nc(){
static char buf[100000],*p1=buf,*p2=buf;
return p1==p2&&(p2=(p1=buf)+fread(buf,1,100000,stdin),p1==p2)?EOF:*p1++;
}
inline int read()
{
register int x=0,f=1;register char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
while(ch>='0'&&ch<='9')x=(x<<3)+(x<<1)+ch-'0',ch=getchar();
return x*f;
}
inline void write(register int x)
{
if(!x)putchar('0');if(x<0)x=-x,putchar('-');
static int sta[20];register int tot=0;
while(x)sta[tot++]=x%10,x/=10;
while(tot)putchar(sta[--tot]+48);
}
inline int power(register int a,register int b)
{
int res=1;
while(b)
{
if(b&1)
res=1ll*res*a%mod;
a=1ll*a*a%mod;
b>>=1;
}
return res;
}
int fac[N+M];
int T,n,m,l,r,ans,a[N],b[N],c[N],t[N];
inline void solve()
{
n=read(),m=read(),l=read(),r=read();
for(register int i=1;i<=n;++i)
a[i]=read();
ans=fac[n+m];
int t1=0,t2=0,top=0;
sort(a+1,a+1+n);
for(register int i=1;i<=n;++i)
if(l<=a[i]&&a[i]<=r)
b[++t1]=a[i];
else
c[++t2]=a[i];
b[t1+1]=c[t2+1]=0;
for(register int i=2,res=1;i<=t2+1;++i)
if(c[i]!=c[i-1])
ans=1ll*ans*power(fac[res],mod-2)%mod,res=1;
else
++res;
for(register int i=2,res=1;i<=t1+1;++i)
if(b[i]!=b[i-1])
t[++top]=res,res=1;
else
++res;
sort(t+1,t+top+1);
int lef=r-l+1-top,nw=0,i=1;
for(;i<=top;++i)
{
if(1ll*lef*(t[i]-nw)<=m)
m-=1ll*lef*(t[i]-nw),nw=t[i],++lef;
else
break;
}
nw+=m/lef;
m%=lef;
ans=1ll*ans*power(power(fac[nw],mod-2),lef-m)%mod;
ans=1ll*ans*power(power(fac[nw+1],mod-2),m)%mod;
for(;i<=top;++i)
ans=1ll*ans*power(fac[t[i]],mod-2)%mod;
write(ans),puts("");
}
int main()
{
fac[0]=1;
for(register int i=1;i<N+M;++i)
fac[i]=1ll*fac[i-1]*i%mod;
T=read();
while(T--)
solve();
return 0;
}T2 Game
According to Ke multiple relationships in order to build a map, give chestnut: 1, 2, 6 even want to side, side 1,2,4 ...... even want to 8, resolve this Ke linear sieve
Ke we to conclude that all should click on the degree to 0, set the number \ (k \) , then
\ [Ans = \ sum_ {i = k} ^ ni \ tbinom {i-1} {k-1} k! (Nk)! \]
\ (\ tbinom {i-1 } {k-1} \) represents the \ (I \) th point is taken as a point of 0 degree, in front of \ (i-1 \) th points have taken \ ( k-1 \) th point of the program number of degree 0, \ (K! \) and \ ((NK)! \) indicates whether the number of degrees in the ordering scheme of the 0
Violence calculate what you can, complexity is \ (O (n) \)
#include <bits/stdc++.h>
#define N 10000005
#define mod 1000000007
#define getchar nc
using namespace std;
inline char nc(){
static char buf[100000],*p1=buf,*p2=buf;
return p1==p2&&(p2=(p1=buf)+fread(buf,1,100000,stdin),p1==p2)?EOF:*p1++;
}
inline int read()
{
register int x=0,f=1;register char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
while(ch>='0'&&ch<='9')x=(x<<3)+(x<<1)+ch-'0',ch=getchar();
return x*f;
}
inline void write(register int x)
{
if(!x)putchar('0');if(x<0)x=-x,putchar('-');
static int sta[20];register int tot=0;
while(x)sta[tot++]=x%10,x/=10;
while(tot)putchar(sta[--tot]+48);
}
inline int power(register int a,register int b)
{
int res=1;
while(b)
{
if(b&1)
res=1ll*res*a%mod;
a=1ll*a*a%mod;
b>>=1;
}
return res;
}
int l,r,n,ans,cnt;
int ispr[N],pr[N],tot,fac[N],invf[N];
inline void init()
{
ispr[1]=pr[1]=1;
if(l==1)
{
cnt=1;
return;
}
for(register int i=2;i<=r;++i)
{
if(!ispr[i])
pr[++tot]=i,cnt+=(i>=l);
for(register int j=1;j<=tot&&pr[j]*i<=r;++j)
{
ispr[pr[j]*i]=1;
if(i<l&&i*pr[j]>=l)
++cnt;
if(i%pr[j]==0)
break;
}
}
}
int main()
{
l=read(),r=read(),n=r-l+1;
init();
fac[0]=1;
for(register int i=1;i<=n;++i)
fac[i]=1ll*fac[i-1]*i%mod;
invf[n]=power(fac[n],mod-2);
for(register int i=n-1;i>=0;--i)
invf[i]=1ll*invf[i+1]*(i+1)%mod;
for(register int i=cnt;i<=n;++i)
ans=(0ll+ans+1ll*i*fac[i-1]%mod*invf[i-cnt]%mod)%mod;
write(1ll*ans*invf[cnt-1]%mod*fac[cnt]%mod*fac[n-cnt]%mod);
return 0;
}T3 guard
Quite magic a question, it must estimate the examination room because of this problem and 200pts retired
Set \ (f [l] [r ] \) represents \ ([l, r] \ ) of the answer,Shifted easily transfer(Read the program will be able to understand), complexity is \ (O (n ^ 2) \)
#include <bits/stdc++.h>
#define N 5005
#define getchar nc
using namespace std;
inline char nc(){
static char buf[100000],*p1=buf,*p2=buf;
return p1==p2&&(p2=(p1=buf)+fread(buf,1,100000,stdin),p1==p2)?EOF:*p1++;
}
inline int read()
{
register int x=0,f=1;register char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
while(ch>='0'&&ch<='9')x=(x<<3)+(x<<1)+ch-'0',ch=getchar();
return x*f;
}
inline void write(register int x)
{
if(!x)putchar('0');if(x<0)x=-x,putchar('-');
static int sta[20];register int tot=0;
while(x)sta[tot++]=x%10,x/=10;
while(tot)putchar(sta[--tot]+48);
}
int n,h[N],see[N][N],f[N][N],ans;
int main()
{
n=read();
for(register int i=1;i<=n;++i)
h[i]=read();
for(register int i=1,l=0;i<=n;++i,l=i-1)
for(double mx=1e9;l>=1;--l)
if(1.0*(h[i]-h[l])/(i-l)<mx)
mx=1.0*(h[i]-h[l])/(i-l),see[l][i]=1;
for(register int r=1;r<=n;++r)
for(register int l=r,s=1,las=l;l>=1;--l)
{
if(see[l][r])
{
if(!see[l+1][r])
s+=min(f[l+1][las],f[l+1][las+1]);
f[l][r]=s;
}
else
{
if(see[l+1][r])
las=l;
f[l][r]=s+min(f[l][las],f[l][las+1]);
}
ans^=f[l][r];
}
write(ans);
return 0;
}