Znzryb's Blog

2025牛客暑期多校训练营2——Bitwise Perfect

Posted on 2025-07-17 Edited on 2026-04-06

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 2026-04-06

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 2026-04-06

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 2026-04-06

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
*/

心路历程（WA，TLE，MLE……）

河南萌新联赛2025第（一）场：河南工业大学——M.无聊的子序列

Posted on 2025-07-16 Edited on 2026-04-06

思路讲解

给定 r 和 s，任何长度至少为(r − 1)(s − 1) + 1 的互异实数序列都包含长度为 r 的单调递增子序列或长度为 s 的单调递减子序列。

$L=(r-1)*(s-1)+1 =2*2+1=5$

我们来分解一下这个定义：

子序列 (Subsequence)：从原序列中，按照原来的顺序，拿掉零个或多个元素后，剩下的元素组成的序列。例如，在序列 {3, 1, 4, 5, 2} 中，{1, 4, 2} 就是一个子序列。
单调递增 (Monotonically Increasing)：序列中的每个数都比它前面的数要大。例如 {2, 5, 8, 10}。
单调递减 (Monotonically Decreasing)：序列中的每个数都比它前面的数要小。例如 {9, 6, 3, 1}。

所以，这个定理告诉你，只要你的序列足够长，你就一定能从中找到一个特定长度的、要么是持续上升、要么是持续下降的子序列。

AC代码

https://ac.nowcoder.com/acm/contest/view-submission?submissionId=78240193

源代码

// Problem: 无聊的子序列
// Contest: NowCoder
// URL: https://ac.nowcoder.com/acm/contest/113664/M
// Memory Limit: 512 MB
// Time Limit: 2000 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)

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 get3(ll x){
	if(x+2>N) return 0;
	if(A[x]<= A[x+1] && A[x+1]<=A[x+2])return 0;
	if(A[x]>=A[x+1] && A[x+1]>=A[x+2]) return 0;
	return 1;
}
ll get4(ll x){
	if(x+3>N) return 0;
	if(A[x]<= A[x+1] && A[x+1]<=A[x+2])return 0;
	if(A[x]>=A[x+1] && A[x+1]>=A[x+2]) return 0;
	if(A[x]<= A[x+1] && A[x+1]<=A[x+3])return 0;
	if(A[x]>= A[x+1] && A[x+1]>=A[x+3])return 0;
	if(A[x]>=A[x+2] && A[x+2]>=A[x+3]) return 0;
	if(A[x]<=A[x+2] && A[x+2]<=A[x+3]) return 0;
	if(A[x+1]>=A[x+2] && A[x+2]>=A[x+3]) return 0;
	if(A[x+1]<=A[x+2] && A[x+2]<=A[x+3]) return 0;
	return 1;
}
inline void __(){
	cin>>N;
	FOR(i,1,N){
		cin>>A[i];
	}
	ll ans=N+N-1;
	FOR(i,1,N){
		ans+=get3(i)+get4(i);
	}
	cout<<ans<<"\n";
}

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

*/