NOIP analog 85 (multi school 18)

preface

It seems that the person described in the background of each topic is a role in a certain film. It seems that it is very tragic (emotional).

Then I only know T1 berries. It's really miserable...

T1 berry conscience

Problem solving ideas

First, the answer is only related to the sum of \ (w \), so the problem becomes to find the size of each group and the number of corresponding schemes for a point.

In the examination room, you want to enumerate the size of the group, and then preprocess the scheme number of \ (n \) numbers divided into \ (m \) non empty sets.

At first, I thought about the partition method, which was obviously wrong. Then I tried DP calculation, and there was \ (f_{i,j}=f_{i-1,j}\times j+f_{i-1,j-1} \)

I've been thinking about the fast power of the matrix. After watching it for a long time, I found that these TM two dimensions, I'm a ghost!!

Only after the test did I know that this is the second kind of Stirling number, and then I found that this thing can calculate the value of a row or a column by NTT. I'm shallow...

The practice of official problem solving is very different. For a pair of \ (u,v \), when they are divided into a group, there will be more contribution of \ (w_u+w_v \).

Then the answer is \ (\ sum \ limits {I = 1} ^ n w_i) \ times (\; {n \ brace K} + (n-1) \ times {n-1 \ brace K} \;) \)

It is found that the following can be directly excluded:

\[\displaystyle{n \brace k}=\frac{1}{k!}\sum_{i=0}^k(-1)^i \binom{k}{i}(k-i)^n \]

Then the linear sieve of \ (\ mathcal{O}(n) \) or the fast power of \ (\ mathcal{O}(nlogn) \).

code

#include<bits/stdc++.h>
#define int long long
#define ull unsigned long long
#define f() cout<<"RP++"<<endl
using namespace std;
inline int read()	
{
	int x=0,f=1;char ch=getchar();
	while(ch>'9'||ch<'0'){if(ch=='-')f=-1;ch=getchar();}
	while(ch>='0'&&ch<='9'){x=(x<<1)+(x<<3)+(ch^48);ch=getchar();}
	return x*f;
}
const int N=1e6+10,M=2e3+10,mod=998244353;
int n,m,ans,base,fac[N],ifac[N] ;
int power(int x,int y,int p=mod)
{
	int temp=1;
	while(y)
	{
		if(y&1) temp=temp*x%p;
		x=x*x%p; y>>=1;
	}
	return temp;
}
int C(int x,int y){return fac[x]*ifac[y]%mod*ifac[x-y]%mod;}
int STL(int x,int y)
{
    int temp=0;
    for(int i=0,bas=1;i<=y;i++,bas=-bas) temp=(temp+bas*C(y,i)*power(y-i,x)%mod+mod)%mod;
    return temp*ifac[y]%mod; 
}
#undef int
int main()
{
	#define int long long
	freopen("ichigo.in","r",stdin); freopen("ichigo.out","w",stdout);
	n=read(); m=read(); fac[0]=ifac[0]=1;
	for(int i=1;i<=n;i++) base=(base+read())%mod,fac[i]=fac[i-1]*i%mod;
	ifac[n]=power(fac[n],mod-2); for(int i=n-1;i>=1;i--) ifac[i]=ifac[i+1]*(i+1)%mod;
	printf("%lld",base*(STL(n,m)%mod+STL(n-1,m)%mod*(n-1)%mod)%mod);
	return 0;
}

I've run out of pears

Problem solving ideas

A direct DP is the shortest time to buy \ (j \) items in \ (I \) stores before \ (f_{i,j} \).

It is found that when \ (a_i \) is not 0, the growth is exponential, so each transfer only needs \ (log \) times.

Then, handle the case where \ (a_i \) is 0, and directly maintain a prefix and binary.

There are some small details in the code implementation.

code

#include<bits/stdc++.h>
#define int long long
#define ull unsigned long long
#define f() cout<<"RP++"<<endl
using namespace std;
inline int read()	
{
	int x=0,f=1;char ch=getchar();
	while(ch>'9'||ch<'0'){if(ch=='-')f=-1;ch=getchar();}
	while(ch>='0'&&ch<='9'){x=(x<<1)+(x<<3)+(ch^48);ch=getchar();}
	return x*f;
}
const int N=2e5+10;
int n,m,ans,pos,lim,f[N],pre[N];
struct Node{int a,b;}s[N];
bool comp(Node x,Node y){x.b++;y.b++;if(x.b*y.a!=y.b*x.a)return x.b*y.a<y.b*x.a;return x.b<y.b;}
#undef int
int main()
{
	#define int long long
	freopen("eriri.in","r",stdin); freopen("eriri.out","w",stdout);
	n=read(); m=read(); lim=(int)log2(m)+1; memset(f,0x3f,sizeof(f));
	for(int i=1;i<=n;i++) s[i].a=read(),s[i].b=read();
	f[0]=0; sort(s+1,s+n+1,comp); pos=n+1;
	for(int i=1;i<=n;i++) if(!s[i].a){pos=i;break;}
	for(int i=1;i<=n;i++) s[i].b+=s[i].a+1,s[i].a++;
	for(int i=pos;i<=n;i++) pre[i]=pre[i-1]+s[i].b;
	for(int i=1;i<pos;i++)
		for(int j=min(i,lim);j>=1;j--)
			if(f[j-1]<=m) f[j]=min(f[j],f[j-1]*s[i].a+s[i].b);
	for(int i=0;i<=min(n,lim);i++)
	{
		if(f[i]>m) continue;
		int l=pos,r=n,temp=-1;
		while(l<=r)
		{
			int mid=(l+r)>>1;
			if(pre[mid]<=m-f[i]) temp=mid,l=mid+1;
			else r=mid-1;
		}
		ans=max(ans,i+((~temp)?temp-pos+1:0));
	}
	printf("%lld",ans); 
	return 0;
}

T3 regiment, but

Problem solving ideas

A wonderful question.

Let \ (p(i) \) represent the number of schemes for \ (i \) rockfill \ (p(i)=(2^n-1)^{\underline{i} \).

Let \ (f(n) \) represent the number of schemes that \ (n \) will fail first. Consider adding a pile of stones equal to the XOR and sum of \ (i-1 \) after \ (i-1 \).

That is \ (p(n-1) \), but if the exclusive or sum of \ (i-1 \) rockfill is already 0, it is obviously illegal and needs to be subtracted.

In another case, the XOR sum of \ (i-2 \) stones is 0, and two new piles of the same stones are added. Obviously, it is also impossible and need to be subtracted.

So there is:

\[f(i)=p(i-1)-f(i-1)-f(i-2)\times(i-1)\times(2^n-i+1) \]

Direct recursion, and then subtract the total number of schemes.

code

#include<bits/stdc++.h>
#define int long long
#define ull unsigned long long
#define f() cout<<"RP++"<<endl
using namespace std;
inline int read()	
{
	int x=0,f=1;char ch=getchar();
	while(ch>'9'||ch<'0'){if(ch=='-')f=-1;ch=getchar();}
	while(ch>='0'&&ch<='9'){x=(x<<1)+(x<<3)+(ch^48);ch=getchar();}
	return x*f;
}
const int N=1e7+10,mod=1e9+7;
int n,p[N],p2[N],f[N];
#undef int
int main()
{
	#define int long long
	freopen("yui.in","r",stdin); freopen("yui.out","w",stdout);
	n=read(); p2[0]=1; for(int i=1;i<=n;i++) p2[i]=p2[i-1]*2%mod;
	p[0]=1; for(int i=1;i<=n;i++) p[i]=p[i-1]*(p2[n]-i)%mod;
	for(int i=3;i<=n;i++) f[i]=(p[i-1]-f[i-1]-(i-1)*f[i-2]%mod*(p2[n]-i+1)%mod+2*mod)%mod;
	printf("%lld",(p[n]-f[n]+mod)%mod);
	return 0;
}

T4 seven negative me

Problem solving ideas

The optimal strategy is that we evenly distribute all values into a complete graph, which can be proved by the adjustment method.

Then, direct enumeration is unacceptable for \ (2^n \). Consider \ (meet\;in\;the\;middle \)

So enumerate the complete graph in the front \ (\ frac{n}{2} \), and then look at its edge connection for the next \ (\ frac{n}{2} \).

After preprocessing, \ (\ frac{n}{2} \) points, the number of complete graphs contained in all point sets.

Then enumerate the point sets of the first \ (\ frac{n}{2} \) points, and then look at their union and combination of the edges of the last \ (\ frac{n}{2} \).

There seems to be a detail wrong in the code, but the wrong point is less than 20. I directly... Combination fist??

code

#include<bits/stdc++.h>
#define int long long
#define ull unsigned long long
#define f() cout<<"RP++"<<endl
using namespace std;
inline int read()	
{
	int x=0,f=1;char ch=getchar();
	while(ch>'9'||ch<'0'){if(ch=='-')f=-1;ch=getchar();}
	while(ch>='0'&&ch<='9'){x=(x<<1)+(x<<3)+(ch^48);ch=getchar();}
	return x*f;
}
const int N=50;
int n,m,m1,m2,all,maxn,e[N],f[1<<20];
#undef int
int main()
{
	#define int long long
	freopen("nanami.in","r",stdin); freopen("nanami.out","w",stdout);
	n=read(); m=read(); all=read(); m1=n>>1; m2=n-m1; if(n<=20) m1=n;
	for(int i=1,x,y;i<=m;i++) x=read(),y=read(),e[x]|=1ll<<y-1,e[y]|=1ll<<x-1;
	for(int sta=1;sta<(1ll<<m1);sta++)
	{
		int sum=__builtin_popcount(sta); if(sum<=maxn) continue;
		for(int i=1;i<=m1;i++) if((sta>>i-1)&1) if((e[i]&sta)!=(sta^(1ll<<i-1))) goto X;
		maxn=max(maxn,sum);X:;
	}
	if(n<=20) printf("%.6lf",(1.0*(maxn*(maxn-1)/2))*(1.0*all)/(1.0*maxn)*(1.0*all)/(1.0*maxn)),exit(0);
	for(int sta=1;sta<(1ll<<m2);sta++)
	{
		int sum=__builtin_popcount(sta);
		for(int i=1;i<=m2;i++) if((sta>>i-1)&1) if(((e[i+m1]>>m1)&sta)!=(sta^(1ll<<i-1))) goto Y;
		f[sta]=sta; maxn=max(maxn,sum); Y:;
	}
	for(int sta=0;sta<(1ll<<m2);sta++)
	{
		int U=(1ll<<m2)-1,sum=__builtin_popcount(sta);
		for(int i=1;i<=m2;i++) if((sta>>i-1)&1) U&=e[i+m1]>>m1;
		if(!U&&sum!=1) continue; 
		for(int i=1;i<=m2;i++)
			if((((sta>>i-1)&1)^1))
				if((((e[i+m1]>>m1)&sta)==sta&&(U&(1ll<<i-1)))){if(__builtin_popcount(f[sta|(1ll<<i-1)])<sum+1) f[sta|(1ll<<i-1)]=f[sta]|(1ll<<i-1);}
				else if(__builtin_popcount(f[sta|(1ll<<i-1)])<sum) f[sta|(1ll<<i-1)]=f[sta];
	}
	for(int sta=1;sta<(1ll<<m1);sta++)
	{
		int U=(1<<m2)-1,sum=__builtin_popcount(sta);
		for(int i=1;i<=m1;i++) if((sta>>i-1)&1) if((e[i]&sta)!=(sta^(1ll<<i-1))) goto Z;
		for(int i=1;i<=m1;i++) if((sta>>i-1)&1) U&=e[i]>>m1;
		maxn=max(maxn,sum+__builtin_popcount(f[U])); Z:;
	}
	printf("%.6lf",(1.0*(maxn*(maxn-1)/2))*(1.0*all)/(1.0*maxn)*(1.0*all)/(1.0*maxn));
	return 0;
}

Posted by hannah415 on Fri, 29 Oct 2021 03:59:20 -0700

Programmer Group

NOIP analog 85 (multi school 18)

preface

T1 berry conscience

Problem solving ideas

code

I've run out of pears

Problem solving ideas

code

T3 regiment, but

Problem solving ideas

code

T4 seven negative me

Problem solving ideas

code

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