Znzryb's Blog

HDU - 5917-Instability

Posted on 2025-05-19 Edited on 2026-02-16

思路讲解

根据 Ramsey定理，大于等于6个的集合，肯定存在一个子集的边都是红色或者都是蓝色，即为团还是为孤立点；

即6 个人中至少存在3人相互认识或者相互不认识。

FOR(i,6,N){		// 根据Ramsey定理的结论
	ans+=C(N,i);
	ans%=mod;
}

然后3-5个点的枚举就可以了

AC代码

https://vjudge.net/solution/60814540

源代码

// Problem: Instability
// Contest: Virtual Judge - HDU
// URL: https://vjudge.net/problem/HDU-5917
// Memory Limit: 64 MB
// Time Limit: 1000 ms
// by znzryb
// 
// Powered by CP Editor (https://cpeditor.org)

#include <bits/stdc++.h>
#define all(vec) vec.begin(),vec.end()
#define CLR(i,a) memset(i,a,sizeof(i))
#define fi first
#define se second
#define pb push_back
#define SZ(a) ((int) a.size())
#define FOR(i, a, b) for (int i = (a); i <= (b); ++i)
#define ROF(i, a, b) for (int i = (a); i >= (b); --i)
#define getFro(vec) (vec.empty()?0:vec.front())
#define getBac(vec) (vec.empty()?0:vec.front())
#define debug(var) cerr << #var <<":"<<var<<"\n";
#define DEBUG(variable) \
    do { \
        std::cerr << #variable << ":"; \
        for (const auto& elem : variable) { \
            std::cerr << elem << " "; \
        } \
        std::cerr << "\n"; \
    } while (0)
#define uniVec(var) \
    do { \
        sort(var.begin(),var.end());\
        var.resize(unique(var.begin(),var.end())-var.begin());\
    } while (0)
#define debugN(var,N) \
	do{ \
		std::cerr<<#var<<":"; \
		FOR(i,0,N){ \
			std::cerr<<var[i]<<" "; \
		} \
		std::cerr<<"\n"; \
	}while(0)
#define debugMap(variable) \
    do { \
        std::cerr << #variable << ":\n"; \
        for (const auto& pair : variable) { \
            std::cerr << "  " << pair.first << " => " << pair.second << "\n"; \
        } \
        std::cerr << std::endl; \
    } while (0)
#define lson(var) (var<<1)
#define rson(var) ((var<<1)+1)

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef __int128 i128;
typedef pair<ll,ll> pll;
typedef array<ll,3> arr;
typedef double DB;
typedef long double LD;
typedef pair<DB,DB> pdd;
typedef pair<ll,bool> plb;
constexpr ll MAXN=static_cast<ll>(1e6)+10,INF=static_cast<ll>(1e18)+3;
constexpr ll mod=static_cast<ll>(1e9)+7;
constexpr double eps=1e-8;

ll N,M,K,T,oT;
bool G[55][55];
/*

*/
ll frac[60],infrac[60];
inline ll C(ll n,ll k){
	ll res=frac[n]*infrac[n-k];
	res%=mod;
	res*=infrac[k];
	res%=mod;
	return res;
}
inline ll binpow(ll a,ll k){
	ll res=1;
	while(k){
		if(k&1) res*=a;
		a*=a;
		a%=mod;
		res%=mod;
		k>>=1;
	}
	return res;
}
inline void solve(){
	cin>>N>>M;
	FOR(i,0,N+3){
		FOR(j,0,N+3){
			G[i][j]=false;
		}
	}
	
	FOR(i,1,M){
		ll u,v;
		cin>>u>>v;
		G[u][v]=true;
		G[v][u]=true;
	}
	ll ans=0;
	FOR(i,6,N){		// 根据Ramsey定理的结论
		ans+=C(N,i);
		ans%=mod;
	}
	FOR(i,1,N){
		FOR(j,i+1,N){
			FOR(k,j+1,N){
				bool alTrue=false;
				// if(G[i][j]==false && G[i][k]==false && G[k][j]==false){
					// ans+=1;
					// alTrue=true;
				// }else if(G[i][j] && G[i][k] && G[j][k]){
					// ans+=1;
					// alTrue=true;
				// }
				auto cal=[&](ll i,ll j,ll k){
					if(G[i][j]==false && G[i][k]==false && G[k][j]==false){
						alTrue=true;
					}
					if(G[i][j] && G[i][k] && G[j][k]){
						alTrue=true;
					}
				};
				cal(i,j,k);
				if(alTrue){
					ans+=1;
				}
				FOR(i4,k+1,N){
					alTrue=false;
					cal(i,j,k);
					cal(i,j,i4);
					cal(i,i4,k);
					cal(i4,j,k);
					if(alTrue){
						ans+=1;
					}
					FOR(i5,i4+1,N){
						alTrue=false;
						cal(i,j,k);
						cal(i,j,i4);
						cal(i,i4,k);
						cal(i4,j,k);
						cal(i,j,i5);
						cal(i,i5,k);
						cal(i5,j,k);
						cal(i5,j,i4);
						cal(i5,i4,k);
						cal(i5,i,i4);
						if(alTrue){
							ans+=1;
						}
					}
				}
			}
		}
	}
	ans%=mod;
	if(ans<0) ans+=mod;
	ll Case=oT-T;
	cout<<"Case #"<<Case<<": "<< ans <<"\n";
}

int main()
{
	ios::sync_with_stdio(false);
	cin.tie(0);cout.tie(0);
	cin>>T;
	oT=T;
	frac[0]=1;
	FOR(i,1,50){
		frac[i]=frac[i-1]*i;
		frac[i]%=mod;
	}
	infrac[0]=1;
	FOR(i,1,50){
		infrac[i]=binpow(frac[i],mod-2);
	}
#ifdef LOCAL
    debugN(frac,50);
    cerr<<"\n";
    debugN(infrac,50);
    debug(C(4,2));
#endif
	while(T--){
		solve();
	}
	// solve();
	return 0;
}
/*
AC https://vjudge.net/solution/60814540
1
7 8
1 2
2 3
3 4
4 5
5 6
6 7
1 7
1 4
ans:Case #1: 53

1
5 6
1 2
2 3
3 4
2 5
4 5
1 5
ans: Case #1: 4
*/

心路历程（WA，TLE，MLE……）

小心 infrac[0]=0;

infrac[0]=1;
FOR(i,1,50){
	infrac[i]=binpow(frac[i],mod-2);
}

HDU - 1205-吃糖果

Posted on 2025-05-19 Edited on 2026-02-16

思路讲解

隔板法

如果说有maxA个挡板，至少要有maxA-1个数才能让隔板不相邻。

AC代码

https://vjudge.net/solution/60783771

源代码

// Problem: 吃糖果
// Contest: Virtual Judge - HDU
// URL: https://vjudge.net/problem/HDU-1205
// Memory Limit: 32 MB
// Time Limit: 3000 ms
// by znzryb
// 
// Powered by CP Editor (https://cpeditor.org)

#include <bits/stdc++.h>
#define all(vec) vec.begin(),vec.end()
#define CLR(i,a) memset(i,a,sizeof(i))
#define fi first
#define se second
#define pb push_back
#define SZ(a) ((int) a.size())
#define FOR(i, a, b) for (int i = (a); i <= (b); ++i)
#define ROF(i, a, b) for (int i = (a); i >= (b); --i)
#define getFro(vec) (vec.empty()?0:vec.front())
#define getBac(vec) (vec.empty()?0:vec.front())
#define debug(var) cerr << #var <<":"<<var<<"\n";
#define DEBUG(variable) \
    do { \
        std::cerr << #variable << ":"; \
        for (const auto& elem : variable) { \
            std::cerr << elem << " "; \
        } \
        std::cerr << "\n"; \
    } while (0)
#define uniVec(var) \
    do { \
        sort(var.begin(),var.end());\
        var.resize(unique(var.begin(),var.end())-var.begin());\
    } while (0)
#define debugN(var,N) \
	do{ \
		std::cerr<<#var<<":"; \
		FOR(i,1,N){ \
			std::cerr<<var[i]<<" "; \
		} \
		std::cerr<<"\n"; \
	}while(0)
#define debugMap(variable) \
    do { \
        std::cerr << #variable << ":\n"; \
        for (const auto& pair : variable) { \
            std::cerr << "  " << pair.first << " => " << pair.second << "\n"; \
        } \
        std::cerr << std::endl; \
    } while (0)
#define lson(var) (var<<1)
#define rson(var) ((var<<1)+1)

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef __int128 i128;
typedef pair<ll,ll> pll;
typedef array<ll,3> arr;
typedef double DB;
typedef long double LD;
typedef pair<DB,DB> pdd;
typedef pair<ll,bool> plb;
constexpr ll MAXN=static_cast<ll>(1e6)+10,INF=static_cast<ll>(1e18)+3;
constexpr ll mod=static_cast<ll>(1e9)+7;
constexpr double eps=1e-8;

ll N,M,K,T,A[MAXN];
/*

*/
inline void solve(){
	cin>>N;
	ll sum=0,maxA=0;
	FOR(i,1,N){
		cin>>A[i];
		sum+=A[i];
		maxA=max(A[i],maxA);
	}
	// 如果说有maxA个挡板，至少要有maxA-1个数才能让隔板不相邻
	if(sum-maxA<maxA-1){
		cout<<"No\n";
	}else{
		cout<<"Yes\n";
	}
}

int main()
{
	ios::sync_with_stdio(false);
	cin.tie(0);cout.tie(0);
	cin>>T;
	while(T--){
		solve();
	}
	// solve();
	return 0;
}
/*
AC
https://vjudge.net/solution/60783771
*/

心路历程（WA，TLE，MLE……）

POJ-2096-Collecting Bugs

Posted on 2025-05-18 Edited on 2026-03-23

思路讲解

ABC-382-E - Expansion Packs

AC代码

其实WA了，不过这种题目比较老，不一定有special judge，可能出现精度问题。

我也稍微对拍了几组，应该没什么太大问题。

源代码

// Problem: Collecting Bugs
// Contest: POJ - Northeastern Europe 2004
// URL: http://poj.org/problem?id=2096
// Memory Limit: 64 MB
// Time Limit: 10000 ms
// by znzryb
// 
// Powered by CP Editor (https://cpeditor.org)

#include <bits/stdc++.h>
#define all(vec) vec.begin(),vec.end()
#define CLR(i,a) memset(i,a,sizeof(i))
#define fi first
#define se second
#define pb push_back
#define SZ(a) ((int) a.size())
#define FOR(i, a, b) for (int i = (a); i <= (b); ++i)
#define ROF(i, a, b) for (int i = (a); i >= (b); --i)
#define getFro(vec) (vec.empty()?0:vec.front())
#define getBac(vec) (vec.empty()?0:vec.front())
#define debug(var) cerr << #var <<":"<<var<<"\n";
#define DEBUG(variable) \
    do { \
        std::cerr << #variable << ":"; \
        for (const auto& elem : variable) { \
            std::cerr << elem << " "; \
        } \
        std::cerr << "\n"; \
    } while (0)
#define uniVec(var) \
    do { \
        sort(var.begin(),var.end());\
        var.resize(unique(var.begin(),var.end())-var.begin());\
    } while (0)
#define debugN(var,N) \
	do{ \
		std::cerr<<#var<<":"; \
		FOR(i,1,N){ \
			std::cerr<<var[i]<<" "; \
		} \
		std::cerr<<"\n"; \
	}while(0)
#define debugMap(variable) \
    do { \
        std::cerr << #variable << ":\n"; \
        for (const auto& pair : variable) { \
            std::cerr << "  " << pair.first << " => " << pair.second << "\n"; \
        } \
        std::cerr << std::endl; \
    } while (0)
#define lson(var) (var<<1)
#define rson(var) ((var<<1)+1)

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef __int128 i128;
typedef pair<ll,ll> pll;
typedef array<ll,3> arr;
typedef double DB;
typedef long double LD;
typedef pair<DB,DB> pdd;
typedef pair<ll,bool> plb;
constexpr ll MAXN=static_cast<ll>(1e6)+10,INF=static_cast<ll>(1e18)+3;
constexpr ll mod=static_cast<ll>(1e9)+7;
constexpr double eps=1e-8;

ll N,M,K,T;
// 已经找到了i种bug且在j个子系统里发现了bug，离完成目标还剩多少天数。
DB dp[1010][1010];
/*

*/
inline void solve(){
	ll s;
	cin>>N>>s;
	DB ns=N*s;
	ROF(i,N,0){
		ROF(j,s,0){
			if(i==N && j==s){
				dp[N][s]=0.0;
			}else{
				// 从i+1转移过来，说明bug种类数量增加了，增长的bug不属于原来有的bug
				dp[i][j]=(dp[i+1][j]*(N-i)*j+dp[i][j+1]*(s-j)*i+
				// dp
				dp[i+1][j+1]*(N-i)*(s-j)+ns) /(ns-i*j);
			}
		}
	}
#ifdef LOCAL
    FOR(i,0,N){
    	FOR(j,0,s){
    		debug(dp[i][j]);
    	}
    }
#endif
	cout<<setprecision(4)<<fixed<<dp[0][0]<<"\n";
}

int main()
{
	ios::sync_with_stdio(false);
	cin.tie(0);cout.tie(0);
	// cin>>T;
	// while(T--){
		// solve();
	// }
	solve();
	return 0;
}
/*
WA http://bailian.openjudge.cn/practice/solution/49203195/

*/

心路历程（WA，TLE，MLE……）

2025金马高校联赛-B战术单元协同训练

Posted on 2025-05-18 Edited on 2026-02-16

思路讲解

AC代码

源代码

心路历程（WA，TLE，MLE……）

ABC-405-E - Fruit Lineup

Posted on 2025-05-18 Edited on 2026-02-16

思路讲解

像这种题目，一般要把一对有关系的留到最后，这样子状态就唯一了。

1
2
3

|----A----|     |--D--|
|-------B------|
           |-----C----|

可以先枚举A的边界，然后再把C填好，B，D之间因为有位置关系，所以状态唯一。

// 显然是要枚举水果的边界
ll ans=0;
FOR(i,a,a+b){
	// 在i-1个空位中选a-1个空位(因为第i个位置已经钦定了A（苹果）)
	ll res=comb.C(i-1,a-1);
	res*=comb.C(N-i,c);
	res%=mod;
	ans+=res;
}

AC代码

https://atcoder.jp/contests/abc405/submissions/66027443

源代码

// Problem: E - Fruit Lineup
// Contest: AtCoder - AtCoder Beginner Contest 405
// URL: https://atcoder.jp/contests/abc405/tasks/abc405_e
// Memory Limit: 1024 MB
// Time Limit: 2000 ms
// by znzryb
// 
// Powered by CP Editor (https://cpeditor.org)

#include <bits/stdc++.h>
#define all(vec) vec.begin(),vec.end()
#define CLR(i,a) memset(i,a,sizeof(i))
#define fi first
#define se second
#define pb push_back
#define SZ(a) ((int) a.size())
#define FOR(i, a, b) for (int i = (a); i <= (b); ++i)
#define ROF(i, a, b) for (int i = (a); i >= (b); --i)
#define getFro(vec) (vec.empty()?0:vec.front())
#define getBac(vec) (vec.empty()?0:vec.front())
#define debug(var) cerr << #var <<":"<<var<<"\n";
#define DEBUG(variable) \
    do { \
        std::cerr << #variable << ":"; \
        for (const auto& elem : variable) { \
            std::cerr << elem << " "; \
        } \
        std::cerr << "\n"; \
    } while (0)
#define uniVec(var) \
    do { \
        sort(var.begin(),var.end());\
        var.resize(unique(var.begin(),var.end())-var.begin());\
    } while (0)
#define debugN(var,N) \
	do{ \
		std::cerr<<#var<<":"; \
		FOR(i,1,N){ \
			std::cerr<<var[i]<<" "; \
		} \
		std::cerr<<"\n"; \
	}while(0)
#define debugMap(variable) \
    do { \
        std::cerr << #variable << ":\n"; \
        for (const auto& pair : variable) { \
            std::cerr << "  " << pair.first << " => " << pair.second << "\n"; \
        } \
        std::cerr << std::endl; \
    } while (0)
#define lson(var) (var<<1)
#define rson(var) ((var<<1)+1)

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef __int128 i128;
typedef pair<ll,ll> pll;
typedef array<ll,3> arr;
typedef double DB;
typedef pair<DB,DB> pdd;
typedef pair<ll,bool> plb;
constexpr ll MAXN=static_cast<ll>(1e6)+10,INF=static_cast<ll>(1e18)+3;
constexpr ll mod=998244353;
constexpr double eps=1e-8;

ll N,M,K,T;
/*

*/
struct Comb{
	vector<ll> frac,infrac;
	int ln=1;
	Comb(int n){
		ln=n;
		frac.resize(n+5,1);
		infrac.resize(n+5,1);
		for(int i=1;i<=ln;++i){
			frac[i]=frac[i-1]*i;
			frac[i]%=mod;
		}
		for(int i=1;i<=ln;++i){
			infrac[i]=binpow(frac[i],mod-2);
		}
	}
	inline ll binpow(ll a,ll k){
		ll res=1;
		while(k){
			if(k&1) res*=a;
			a*=a;
			a%=mod;
			res%=mod;
			k>>=1;
		}
		return res;
	}
	inline ll C(ll n,ll k){
		ll res=frac[n]*infrac[n-k];
		res%=mod;
		res*=infrac[k];
		res%=mod;
		return res;
	}
	inline ll P(ll n,ll k){
		ll res=frac[n]*infrac[n-k];
		res%=mod;
		return res;
	}
};
inline void solve(){
	ll a,b,c,d;
	cin>>a>>b>>c>>d;
	N=a+b+c+d;
	Comb comb(a+b+c+d);
	// 显然是要枚举水果的边界
	ll ans=0;
	FOR(i,a,a+b){
		// 在i-1个空位中选a-1个空位(因为第i个位置已经钦定了A（苹果）)
		ll res=comb.C(i-1,a-1);
		res*=comb.C(N-i,c);
		res%=mod;
		ans+=res;
	}
	ans%=mod;
	if(ans<0) ans+=mod;
	cout<<ans<<"\n";
}

int main()
{
	ios::sync_with_stdio(false);
	cin.tie(0);cout.tie(0);
	// cin>>T;
	// while(T--){
		// solve();
	// }
	solve();
	return 0;
}
/*
AC
https://atcoder.jp/contests/abc405/submissions/66027443
*/