Znzryb's Blog

2025牛客暑期多校训练营2——Love Wins All（不要把问题想复杂了）（给的是排列）

Posted on 2025-07-18 Edited on 2026-02-17

思路讲解

一个图中，如果所有节点的入度和出度都正好是1，那么这个图必然是由若干个不相交的环组成的。

题目的排列给出保证了所有点的出度和入度都为1。

可以使用并查集维护。特别需要注意的就是siz==2的情况，虽然cnt不统计，但是后续计算当替罪羊的时候要统计。

// 只可能偶数个奇数环
if(cnt[1]==2){
	// 如果两个奇数环，那么两个不匹配点一定从两个奇数环出
	ans=binpow(2,cnt[0])*mul[1];
	ans%=mod;
}else if(cnt[1]>=4){	// 奇数环数量大于4，必然无法组成合法对
	cout<<0<<"\n";
	return;
}else{
	vis.assign(N+5,0);
	FOR(i,1,N){
		if(vis[fa.find(i)]) continue;
		vis[fa.find(i)]=true;
		ll siz=fa.size(i);
		if(siz==2){    // 这里一定不要遗漏
			ans+=binpow(2,cnt[0]);
			ans%=mod;
		}else{
			// 选择一黑一白，图的剩余部分必然可以完美匹配（注意一黑一白不一定相邻）
			ll t=siz*siz%mod*binpow(4,mod-2);t%=mod;
			ans+=binpow(2,cnt[0]-1)*t;
			ans%=mod;
		}
	}
}

AC代码

https://ac.nowcoder.com/acm/contest/view-submission?submissionId=78310781

源代码

// Problem: Love Wins All
// Contest: NowCoder
// URL: https://ac.nowcoder.com/acm/contest/108299/L
// Memory Limit: 1024 MB
// Time Limit: 4000 ms
// 
// Powered by CP Editor (https://cpeditor.org)

#include <bits/stdc++.h>
#define all(vec) vec.begin(),vec.end()
#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 debug(var) cerr << #var <<":"<<var<<"\n";
#define lson(var) (var<<1)
#define rson(var) ((var<<1)+1)
#define V vector

using namespace std;

using  ll = long long;using ull = unsigned long long;
using DB=double;using LD=long double;
// using i128 = __int128;
using pdd = pair<DB,DB>;using plb = pair<ll,bool>;
using pll = pair<ll,ll>;using vll=vector<ll>;using vb=vector<bool>;
using arr3 = array<ll,3> ;using arr2 = array<ll,2>;
constexpr ll MAXN=static_cast<ll>(5e5)+10,INF=static_cast<ll>(1e17)+9; // 1e18+9也是素数
constexpr ll MAXM=(ll)1e6+10;constexpr ll MAXV=(ll)1e6+10;
constexpr ll mod=998244353;
constexpr double eps=1e-8;const double pi=acos(-1.0);

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

*/
struct DSU { // 并查集
    vector < int > fa, siz;
    int ln;
    DSU(int n) {
        ln = n;
        fa.resize(ln + 5, 0), siz.resize(ln + 5, 0);
        FOR(i, 1, n) {
            fa[i] = i;
            siz[i] = 1;
        }
    }
    int find(int x) {
        while (fa[x] != x) {
            x = fa[x];
            fa[x] = fa[fa[x]];
        }
        return x;
    }
    bool same(int x, int y) {
        return find(x) == find(y);
    }
    bool merge(int x, int y) {
        x = find(x);
        y = find(y);
        if (x == y) {
            return false;
        }
        siz[x] += siz[y];
        fa[y] = x;
        return true;
    }
    int size(int x) {
        return siz[find(x)];
    }
    int countFa() {
        int res = 0;
        FOR(i, 1, ln) {
            if (fa[i] == i) ++res;
        }
        return res;
    }
};
ll binpow(ll a,ll k){	// 迭代快速幂
	ll res=1;
	a%=mod;		// 防止a*a溢出
	while(k>0){
		if((k&1)==1) res=a*res%mod;
		a=a*a%mod;
		k>>=1;
	}
	return res;
}
inline void __(){
	cin>>N;
	DSU fa(N);
	FOR(i,1,N){
		cin>>A[i];
		fa.merge(i,A[i]);
	}
	vb vis(N+5,0);
	vll cnt(3,0),mul(3,1);
	FOR(i,1,N){
		if(vis[fa.find(i)]) continue;
		vis[fa.find(i)]=true;
		ll siz=fa.size(i);
		// if(siz==1) continue;
		if(siz==2) continue;
		cnt[siz%2]++;
		if(siz%2) mul[siz%2]*=siz;
	}
	ll ans=0;
	// 只可能偶数个奇数环
	if(cnt[1]==2){
		// 如果两个奇数环，那么两个不匹配点一定从两个奇数环出
		ans=binpow(2,cnt[0])*mul[1];
		ans%=mod;
	}else if(cnt[1]>=4){	// 奇数环数量大于4，必然无法组成合法对
		cout<<0<<"\n";
		return;
	}else{
		vis.assign(N+5,0);
		FOR(i,1,N){
			if(vis[fa.find(i)]) continue;
			vis[fa.find(i)]=true;
			ll siz=fa.size(i);
			if(siz==2){
				ans+=binpow(2,cnt[0]);
				ans%=mod;
			}else{
				// 选择一黑一白，图的剩余部分必然可以完美匹配（注意一黑一白不一定相邻）
				ll t=siz*siz%mod*binpow(4,mod-2);t%=mod;
				ans+=binpow(2,cnt[0]-1)*t;
				ans%=mod;
			}
		}
	}
	ans%=mod;
	if(ans<0) ans+=mod;
	cout<<ans<<"\n";
}

int main()
{
	ios::sync_with_stdio(false);
	cin.tie(0);cout.tie(0);
	
	cin>>lT;
	while(lT--)
		__();
	return 0;
}
/*

*/

心路历程（WA，TLE，MLE……）

2025牛客暑期多校训练营2——Bitwise Perfect

Posted on 2025-07-17 Edited on 2026-02-17

思路讲解

我们在想问题的时候，只要确定我们检查的数对其他数有没有增加作用，不用太想。

AC代码

https://ac.nowcoder.com/acm/contest/view-submission?submissionId=78290963

源代码

// Problem: Bitwise Perfect
// Contest: NowCoder
// URL: https://ac.nowcoder.com/acm/contest/108299/B
// Memory Limit: 1024 MB
// Time Limit: 4000 ms
// 
// Powered by CP Editor (https://cpeditor.org)

#include <bits/stdc++.h>
#define all(vec) vec.begin(),vec.end()
#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 debug(var) cerr << #var <<":"<<var<<"\n";
#define lson(var) (var<<1)
#define rson(var) ((var<<1)+1)
#define V vector

using namespace std;

using  ll = long long;using ull = unsigned long long;
using DB=double;using LD=long double;
// using i128 = __int128;
using pdd = pair<DB,DB>;using plb = pair<ll,bool>;
using pll = pair<ll,ll>;using vll=vector<ll>;
using arr3 = array<ll,3> ;using arr2 = array<ll,2>;
constexpr ll MAXN=static_cast<ll>(1e6)+10,INF=static_cast<ll>(1e17)+9; // 1e18+9也是素数
constexpr ll MAXM=(ll)1e6+10;constexpr ll MAXV=(ll)1e6+10;
constexpr ll mod=static_cast<ll>(1e9)+7;
constexpr double eps=1e-8;const double pi=acos(-1.0);

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

*/

inline void __(){
	cin>>N;
	// vector<vector<ll> > bits(N+5,vll(N+5,0));
	vll bits(76,0);
	FOR(i,1,N){
		cin>>A[i];
		bitset<70> bi(A[i]);
		FOR(j,0,67){
			if(bi[j]) bits[j]++;
		}
	}
	FOR(i,1,N){
		ll t=__lg(A[i]);
		if(bits[t]>1){
			cout<<"NO\n";
			return;
		}
	}
	cout<<"YES\n";
}

int main()
{
	ios::sync_with_stdio(false);
	cin.tie(0);cout.tie(0);
	
	cin>>lT;
	while(lT--)
		__();
	return 0;
}
/*
AC
https://ac.nowcoder.com/acm/contest/view-submission?submissionId=78290963
*/

心路历程（WA，TLE，MLE……）

2025牛客暑期多校训练营2——Another Day of Sun（01串）

Posted on 2025-07-17 Edited on 2026-03-12

思路讲解

	ll t=binpow(2,tmp,mod),tt=binpow(2,mod-2,mod),tttt=binpow(4,mod-2,mod);
	ll ans=0;
	A[0]=0;
	FOR(i,0,N-1){
		if(A[i]==0 && A[i+1]==1){
			ans+=t;
			ans%=mod;
		}else if(A[i]==-1 && A[i+1]==-1){
			ans+=t*tttt;
			ans%=mod;
		}else if(A[i]==0 && A[i+1]==-1){
			ans+=t*tt;
			ans%=mod;
		}else if(A[i]==-1 && A[i+1]==1){
			ans+=t*tt;
			ans%=mod;
		}
#ifdef LOCAL
    debug(i);debug(ans);
#endif
	}

AC代码

https://ac.nowcoder.com/acm/contest/view-submission?submissionId=78289781

源代码

// Problem: Another Day of Sun
// Contest: NowCoder
// URL: https://ac.nowcoder.com/acm/contest/108299/A
// Memory Limit: 1024 MB
// Time Limit: 4000 ms
// 
// Powered by CP Editor (https://cpeditor.org)

#include <bits/stdc++.h>
#define all(vec) vec.begin(),vec.end()
#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 debug(var) cerr << #var <<":"<<var<<"\n";
#define lson(var) (var<<1)
#define rson(var) ((var<<1)+1)
#define V vector

using namespace std;

using  ll = long long;using ull = unsigned long long;
using DB=double;using LD=long double;
// using i128 = __int128;
using pdd = pair<DB,DB>;using plb = pair<ll,bool>;
using pll = pair<ll,ll>;using vll=vector<ll>;
using arr3 = array<ll,3> ;using arr2 = array<ll,2>;
constexpr ll MAXN=static_cast<ll>(5e5)+10,INF=static_cast<ll>(1e17)+9; // 1e18+9也是素数
constexpr ll MAXM=(ll)5e5+10;constexpr ll MAXV=(ll)1e6+10;
constexpr ll mod=998244353;
constexpr double eps=1e-8;const double pi=acos(-1.0);

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

*/
ll binpow(ll a,ll k,const ll &p){	// 迭代快速幂
	ll res=1;
	a%=mod;		// 防止a*a溢出
	while(k>0){
		if((k&1)==1) res=a*res%p;
		a=a*a%p;
		k>>=1;
	}
	return res;
}
inline void __(){
	cin>>N;
	ll cnt=0,tmp=0;
	FOR(i,1,N){
		cin>>A[i];
		if(A[i]==1){
			++cnt;
		}else if(A[i]==-1){
			++tmp;
		}
	}
	ll t=binpow(2,tmp,mod),tt=binpow(2,mod-2,mod),tttt=binpow(4,mod-2,mod);
	ll ans=0;
	A[0]=0;
	FOR(i,0,N-1){
		if(A[i]==0 && A[i+1]==1){
			ans+=t;
			ans%=mod;
		}else if(A[i]==-1 && A[i+1]==-1){
			ans+=t*tttt;
			ans%=mod;
		}else if(A[i]==0 && A[i+1]==-1){
			ans+=t*tt;
			ans%=mod;
		}else if(A[i]==-1 && A[i+1]==1){
			ans+=t*tt;
			ans%=mod;
		}
#ifdef LOCAL
    debug(i);debug(ans);
#endif
	}
	ans%=mod;
	if(ans<0) ans+=mod;
	cout<<ans<<"\n";
}

int main()
{
	ios::sync_with_stdio(false);
	cin.tie(0);
	//cout.tie(0);
	cin>>lT;
	while(lT--)
		__();
	return 0;
}
/*
AC
https://ac.nowcoder.com/acm/contest/view-submission?submissionId=78289781
*/

心路历程（WA，TLE，MLE……）

START195B——Game (Hard)（分治搜索）

Posted on 2025-07-17 Edited on 2026-03-19

思路讲解

	// 随着钱数的增加，进行的二操作总是在增加的
	function<void(ll,ll,ll,ll)> div=[&](ll k,ll kk,ll p,ll pp)->void{
		if(k>kk){
			return;
		}
		ll mid=k+kk>>1;
		ll ans=Sum[min(N,mid)];ll mp=0;
		FOR(i,p,min(mid/2,pp)){
			ll kejia=mid-i*2;
			kejia+=i;
			kejia=min(kejia,N);
			ll lans=(kejia-1+kejia-i)*i/2;
			lans+=Sum[kejia];
			if(lans>ans){
				ans=lans;
				mp=i;
			}
		}
		Ans[mid]=ans;
#ifdef LOCAL
    debug(mid);
    debug(ans);
    debug(pp);
#endif
		div(k,mid-1,p,mp);
		div(mid+1,kk,mp,pp);
	};

AC代码

https://www.codechef.com/viewsolution/1173480410

源代码

// Problem: Game (Hard)
// Contest: CodeChef - START195B
// URL: https://www.codechef.com/START195B/problems/GAMEHD
// Memory Limit: 256 MB
// Time Limit: 1000 ms
// 
// Powered by CP Editor (https://cpeditor.org)

#include <bits/stdc++.h>
#define all(vec) vec.begin(),vec.end()
#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 debug(var) cerr << #var <<":"<<var<<"\n";
#define lson(var) (var<<1)
#define rson(var) ((var<<1)+1)
#define V vector

using namespace std;

using  ll = long long;using ull = unsigned long long;
using DB=double;using LD=long double;
// using i128 = __int128;
using pdd = pair<DB,DB>;using plb = pair<ll,bool>;
using pll = pair<ll,ll>;using vll=vector<ll>;
using arr3 = array<ll,3> ;using arr2 = array<ll,2>;
constexpr ll MAXN=static_cast<ll>(1e6)+10,INF=static_cast<ll>(1e17)+9; // 1e18+9也是素数
constexpr ll MAXM=(ll)1e6+10;constexpr ll MAXV=(ll)1e6+10;
constexpr ll mod=static_cast<ll>(1e9)+7;
constexpr double eps=1e-8;const double pi=acos(-1.0);

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

*/
ll Sum[MAXN];
inline void __(){
	cin>>N;
	FOR(i,1,N){
		cin>>A[i];
	}
	vll sa(A+1,A+1+N);
	sort(all(sa),greater<>());
	FOR(i,1,N){
		Sum[i]=Sum[i-1]+sa[i-1];
	}
	vll Ans(2*N+5,0);
	// p is to get p Ai
	// 随着钱数的增加，进行的二操作总是在增加的
	function<void(ll,ll,ll,ll)> div=[&](ll k,ll kk,ll p,ll pp)->void{
		if(k>kk){
			return;
		}
		ll mid=k+kk>>1;
		ll ans=Sum[min(N,mid)];ll mp=0;
		FOR(i,p,min(mid/2,pp)){
			ll kejia=mid-i*2;
			kejia+=i;
			kejia=min(kejia,N);
			ll lans=(kejia-1+kejia-i)*i/2;
			lans+=Sum[kejia];
			if(lans>ans){
				ans=lans;
				mp=i;
			}
		}
		Ans[mid]=ans;
#ifdef LOCAL
    debug(mid);
    debug(ans);
    debug(pp);
#endif
		div(k,mid-1,p,mp);
		div(mid+1,kk,mp,pp);
	};
	div(1,2*N,1,N);
	FOR(i,1,2*N){
		cout<<Ans[i]<<" ";
	}
	cout<<"\n";
}

int main()
{
	ios::sync_with_stdio(false);
	cin.tie(0);cout.tie(0);
	
	cin>>lT;
	while(lT--)
		__();
	return 0;
}
/*
AC
https://www.codechef.com/viewsolution/1173480410
*/

心路历程（WA，TLE，MLE……）

START195B——String Deletions（注意前导一致和后导一致）（01字符串）

Posted on 2025-07-17 Edited on 2026-02-17

思路讲解

可以看提交链接

AC代码

https://www.codechef.com/viewsolution/1173432221

源代码

// Problem: String Deletions
// Contest: CodeChef - START195B
// URL: https://www.codechef.com/START195B/problems/STRDEL7
// Memory Limit: 256 MB
// Time Limit: 1000 ms
// 
// Powered by CP Editor (https://cpeditor.org)

#include <bits/stdc++.h>
#define all(vec) vec.begin(),vec.end()
#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 debug(var) cerr << #var <<":"<<var<<"\n";
#define lson(var) (var<<1)
#define rson(var) ((var<<1)+1)
#define V vector

using namespace std;

using  ll = long long;using ull = unsigned long long;
using DB=double;using LD=long double;
// using i128 = __int128;
using pdd = pair<DB,DB>;using plb = pair<ll,bool>;
using pll = pair<ll,ll>;using vll=vector<ll>;
using arr3 = array<ll,3> ;using arr2 = array<ll,2>;
constexpr ll MAXN=static_cast<ll>(1e6)+10,INF=static_cast<ll>(1e17)+9; // 1e18+9也是素数
constexpr ll MAXM=(ll)1e6+10;constexpr ll MAXV=(ll)1e6+10;
constexpr ll mod=static_cast<ll>(1e9)+7;
constexpr double eps=1e-8;const double pi=acos(-1.0);

ll N,M,Q,X,K,T,lT;
/*

*/
char S[MAXN];
inline void __(){
	cin>>N;
	
	FOR(i,1,N){
		cin>>S[i];
		// ++cnt[S[i]-'0'];
	}
	vector<vector<ll> > g(4);
	vll wp;
	ll dec=0;
	FOR(i,1,N){
		if(g[S[i]-'0'].empty() || i-g[S[i]-'0'].back()>1){
			wp.pb(i);
		}else{
			++dec;
// #ifdef LOCAL
    // debug(i);
    // debug(i-g[S[i]-'0'].back());
    // debug(S[i]);
    // debug(g[S[i]-'0'].back());
// #endif
		}
		g[S[i]-'0'].pb(i);
	}
	if(SZ(g[0])==N || SZ(g[1])==N){
		cout<<N<<"\n";
		return;
	}
	ll ans=N-dec;
	if(SZ(wp)>=3){
		vll cnt(4,0);
		ll lans=0;
		if(S[wp.back()]!=S[wp.front()]){
			lans=SZ(wp)-3;
		}else{
			lans=SZ(wp)-2;
		}
		ans-=lans;
	}
	ROF(i,N,1){
		if(S[i]==S[N]){
			++ans;
		}else{
			break;
		}
	}
	FOR(i,1,N){
		if(S[i]==S[1]){
			++ans;
		}else{
			break;
		}
	}
	ans-=1;
	ans-=1;
	cout<<ans<<"\n";
}

int main()
{
	ios::sync_with_stdio(false);
	cin.tie(0);cout.tie(0);
	
	cin>>lT;
	while(lT--)
		__();
	return 0;
}
/*
AC
https://www.codechef.com/viewsolution/1173432221
1
5
11010
4

1
5
01010

ans:2
*/