Znzryb's Blog

2025“钉耙编程”中国大学生算法设计暑期联赛（2）——子集

Posted on 2025-08-01 Edited on 2026-04-06

思路讲解

这个子集确实还可以，不是很难，就是卡一下这个折半搜索01树。

你用线性基做基本不会被卡。

AC代码

源代码

// by Gospel_rock
// from template_cpp
#include <bits/stdc++.h>
#define all(vec) vec.begin(),vec.end()
#define pb push_back
#define EB emplace_back
#define EM emplace
#define FI first
#define SE second
#define pct __builtin_popcountll
#define ctz __builtin_ctzll
#define SZ(a) ((long long) a.size())
#define FOR(i, a, b) for (long long i = (a); i <= (b); ++i)
#define ROF(i, a, b) for (long long i = (a); i >= (b); --i)
#define debug(var) cerr << #var <<":"<<var<<"\n";
#define cend cerr<<"\n-----------\n"
#define lson(var) (var<<1)
#define rson(var) ((var<<1)+1)
#define fsp(x) fixed<<setprecision(x)

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 vdb = vector<DB>;using vc=vector<char>;
using tll = tuple<ll,ll,ll>;
using vpll = vector<pll>;using vtll = vector<tll>;
using arr3 = array<ll,3> ;using arr2 = array<ll,2>;
constexpr ll MAXN=(ll)1e6+10,INF=(ll)1e17+9; // 1e18+9也是素数
constexpr ll mod=(ll)1e9+7; // 998244353;
constexpr double eps=1e-8;const double pi=acos(-1.0);
ll N,M,Q,X,K,T,lT;

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

*/
// https://www.luogu.com.cn/record/228289095 基础模版
// 检测合并功能 https://acm.hdu.edu.cn/contest/view-code?cid=1173&rid=23716
struct LB{
    using ll=long long;
    ll n;vector<ll> b;
    LB(ll n=62){
        this->n=n;
        b.assign(n+5,0);
    }
    void insert(ll x){
        for(int i=n;i>=0;--i){
            if(x>>i&1){
                if(b[i]){
                    x^=b[i];
                }else{
                    b[i]=x;
                    return;
                }
            }
        }
    }
    ll findmx(){
        ll res=0;
        for(int i=n;i>=0;--i){
            if((res^b[i]) > res){
                res^=b[i];
            }
        }
        return res;
    }
    void merge(const LB & t){
        for(int i=n;i>=0;--i){
            if(t.b[i]){
                insert(t.b[i]);
            }
        }
    }
};
inline void __(){
	cin>>N;
    vll A(N);
	FOR(i,1,N){
        cin>>A[i-1];
    }
    ll ans=0;
    function <void(ll,LB)> dfs=[&](ll idx,LB lb)->void{
        if(idx >= N) {
            ans=max(ans,lb.findmx());
            return;
        }
        LB llb=lb;lb.insert(A[idx]);llb.insert(A[idx]);
        dfs(idx+2,lb);
        dfs(idx+3,llb);
    };
    LB lb;
    dfs(0,lb);
    cout<<ans<<"\n";
}
/*
AC
https://acm.hdu.edu.cn/contest/view-code?cid=1173&rid=23716
*/

心路历程（WA，TLE，MLE……）

3630. 划分数组得到最大异或运算和与运算之和

Posted on 2025-08-01 Edited on 2026-04-06

思路讲解

【线性基【力扣周赛 460】】 https://www.bilibili.com/video/BV1pm8vzAEXx/?share_source=copy_web&vd_source=6ca0bc05e7d6f39b07c1afd464edae37

AC代码

https://leetcode.cn/problems/partition-array-for-maximum-xor-and-and/submissions/649067731/

源代码

class Solution {
public:
    using ll =long long ;
    struct LB{
        using ll =long long ;
        ll n;vector<ll> b;
        LB(ll n=62){
            this->n=n;
            b.assign(n+5,0);
        }
        void insert(ll x){
            for(int i=n;i>=0;--i){
                if(x>>i&1){
                    if(b[i]){
                        x^=b[i];
                    }else{
                        b[i]=x;
                        return;
                    }
                }
            }
        }
        ll findmx(){
            ll res=0;
            for(int i=n;i>=0;--i){
                if((res^b[i]) > res){
                    res^=b[i];
                }
            }
            return res;
        }
    };
    long long maximizeXorAndXor(vector<int>& nums) {
        ll N=nums.size();
        ll ans=0;
        // 暴力枚举是否是子集B
        for(int s=0;s<(1<<N);++s){
            bitset<22> bi(s);
            ll ands=-1,xors=0;
            for(int i=0;i<N;++i){
                if(bi[i]){
                    // 是子集B，计算与的和
                    ands&=nums[i];
                }else{
                    xors^=nums[i];
                }
            }
            if(ands==-1) ands=0;
            ll lans=ands+xors;
            // 出现奇数次的1在这个xors里面体现为1
            LB lb;
            for(int i=0;i<N;++i){
                if(!bi[i]) lb.insert(nums[i]&(~xors));
            }
            lans+=lb.findmx()*2;
            ans=max(ans,lans);
        }
        return ans;
    }
};

心路历程（WA，TLE，MLE……）

2025“钉耙编程”中国大学生算法设计暑期联赛（5）——"合理"避税

Posted on 2025-08-01 Edited on 2026-04-06

思路讲解

核心就是知道可以发多少个月，就可以知道总共可以发多少个人次，一个人最多可以发多少钱。

auto check=[&](ll mid){
    // 我们一共可以给mid*P个人次发钱。
    ll chance=mid*P;
    vpll rem;
    ll ans=0;
    FOR(i,1,N){
        ll a=A[i-1];
        ll ci=min({a/K,mid,chance});
        chance-=ci;
        ans+=ci*K;
        ll r=a-K*ci;
        if(r>0) rem.EB(r,mid-ci);
    }
    if(chance<=0){
        if(ans>=M){
            return true;
        }else{
            return false;
        }
    }
    sort(all(rem),greater<>());
    for(auto [v,ci]:rem){
        if(chance==0) break;
        if(ci>0) ans+=v,chance--;
    }
    if(ans>=M){
        return true;
    }else{
        return false;
    }
};

AC代码

https://acm.hdu.edu.cn/contest/view-code?cid=1176&rid=8395

源代码

// by Gospel_rock
// from template_cpp
#include <bits/stdc++.h>
#define all(vec) vec.begin(),vec.end()
#define pb push_back
#define EB emplace_back
#define EM emplace
#define FI first
#define SE second
#define pct __builtin_popcountll
#define ctz __builtin_ctzll
#define SZ(a) ((int) a.size())
#define FOR(i, a, b) for (long long i = (a); i <= (b); ++i)
#define ROF(i, a, b) for (long long i = (a); i >= (b); --i)
#define debug(var) cerr << #var <<":"<<var<<"\n";
#define cend cerr<<"\n-----------\n"
#define lson(var) (var<<1)
#define rson(var) ((var<<1)+1)
#define fsp(x) fixed<<setprecision(x)

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 vdb = vector<DB>;using vc=vector<char>;
using tll = tuple<ll,ll,ll>;
using vpll = vector<pll>;using vtll = vector<tll>;
using arr3 = array<ll,3> ;using arr2 = array<ll,2>;
constexpr ll MAXN=(ll)1e6+10,INF=(ll)1e17+9; // 1e18+9也是素数
constexpr ll mod=(ll)1e9+7; // 998244353;
constexpr double eps=1e-8;const double pi=acos(-1.0);
ll N,M,Q,X,K,T,lT,P;

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

*/
inline void __(){
	cin>>N>>M>>K>>P;
    vll A(N);
	FOR(i,1,N){
        cin>>A[i-1];
    }
    sort(all(A),greater<>());
    auto check=[&](ll mid){
        // 我们一共可以给mid*P个人次发钱。
        ll chance=mid*P;
        vpll rem;
        ll ans=0;
        FOR(i,1,N){
            ll a=A[i-1];
            ll ci=min({a/K,mid,chance});
            chance-=ci;
            ans+=ci*K;
            ll r=a-K*ci;
            if(r>0) rem.EB(r,mid-ci);
        }
        if(chance<=0){
            if(ans>=M){
                return true;
            }else{
                return false;
            }
        }
        sort(all(rem),greater<>());
        for(auto [v,ci]:rem){
            if(chance==0) break;
            if(ci>0) ans+=v,chance--;
        }
        if(ans>=M){
            return true;
        }else{
            return false;
        }
    };
    ll low=1,hig=M/K+(ll)1e4;
    while(low<hig){
        ll mid=low+hig>>1;
        if(check(mid)){
            hig=mid;
        }else{
            low=mid+1;
        }
    }
    cout<<low<<"\n";
}
/*
AC
https://acm.hdu.edu.cn/contest/view-code?cid=1176&rid=8395
*/

心路历程（WA，TLE，MLE……）

2025“钉耙编程”中国大学生算法设计暑期联赛（5）——四角洲行动（乍看起来是背包，但是可以枚举解决）

Posted on 2025-08-01 Edited on 2026-04-06

思路讲解

很复杂？很难？其实枚举就好了。

AC代码

https://acm.hdu.edu.cn/contest/view-code?cid=1176&rid=11903

源代码

// by Gospel_rock
// from template_cpp
#include <bits/stdc++.h>
#define all(vec) vec.begin(),vec.end()
#define pb push_back
#define EB emplace_back
#define EM emplace
#define FI first
#define SE second
#define pct __builtin_popcountll
#define ctz __builtin_ctzll
#define SZ(a) ((long long) a.size())
#define FOR(i, a, b) for (long long i = (a); i <= (b); ++i)
#define ROF(i, a, b) for (long long i = (a); i >= (b); --i)
#define debug(var) cerr << #var <<":"<<var<<"\n";
#define cend cerr<<"\n-----------\n"
#define lson(var) (var<<1)
#define rson(var) ((var<<1)+1)
#define fsp(x) fixed<<setprecision(x)

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 vdb = vector<DB>;using vc=vector<char>;
using tll = tuple<ll,ll,ll>;
using vpll = vector<pll>;using vtll = vector<tll>;
using arr3 = array<ll,3> ;using arr2 = array<ll,2>;
constexpr ll MAXN=(ll)1e6+10,INF=(ll)1e17+9; // 1e18+9也是素数
constexpr ll mod=(ll)1e9+7; // 998244353;
constexpr double eps=1e-8;const double pi=acos(-1.0);
ll N,M,Q,X,K,T,lT;

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

*/
// ceil_div
long long chu(long long x, long long y) {
    if (x * y < 0) return x / y;
    if (x % y == 0) return x / y;
    return x / y + 1;
}

inline void __(){
    ll a,b,c,d;
	cin>>a>>b>>c>>d;
    vector<vll> item(6);
    vector<vll> pre(6);
    FOR(i,1,4){
        ll n;
        cin>>n;
        pre[i].assign(n+5,0);
        FOR(j,1,n){
            ll e;
            cin>>e;
            item[i].pb(e);
        }
        sort(all(item[i]),greater<>());
        FOR(j,1,n){
            pre[i][j]=pre[i][j-1]+item[i][j-1];
        }
    }
    ll ans=0;
    // 先枚举大的，再枚举小的（大的受到的限制多）
    FOR(dd,0,SZ(item[4])){
        if(dd>d) break; 
        FOR(cc,0,SZ(item[3])){
            if(cc>c) break;
            ll lans=0;
            lans+=pre[3][cc];
            lans+=pre[4][dd];
            ll jichu=lans;
            if(SZ(item[1])>a){
                ll ra=SZ(item[1])-a;
                FOR(aa,0,ra){
                    ll rc=c-cc,rd=d-dd;
                    lans=jichu;
                    if(aa>rc*3+rd*4+b*2) break;
                    // 1 的量不要忘记算进去
                    lans+=pre[1][aa+a];
                    if(aa<=rc){
                        lans+=pre[2][min(SZ(item[2]),b+rc+rd*2)];
                        ans=max(lans,ans);
                        continue;
                    }
                    if(aa<=rc*3){
                        ll t=aa-rc;
                        t=chu(t,2);     // 向上取整除法
                        lans+=pre[2][min(SZ(item[2]),b+rc-t+rd*2)];
                        ans=max(lans,ans);
                        continue;
                    }
                    if(aa<=rc*3+rd*2){
                        ll t=aa-rc*3;
                        t=chu(t,2);     // 向上取整除法
                        lans+=pre[2][min(SZ(item[2]),b+(rd-t)*2+t)];
                        ans=max(lans,ans);
                        continue;
                    }
                    if(aa<=rc*3+rd*4){
                        ll t=aa-rc*3-rd*2;
                        t=chu(t,2);     // 向上取整除法
                        lans+=pre[2][min(SZ(item[2]),b+(rd-t))];
                        ans=max(lans,ans);
                        continue;
                    }
                    ll t=aa-rc*3-rd*4;
                    t=chu(t,2);     // 向上取整除法
                    lans+=pre[2][min(SZ(item[2]),b-t)];
                    ans=max(lans,ans);
                }
            }else{  // 没有剩余的1*1节点了
                ll rc=c-cc,rd=d-dd;
                lans+=pre[1][min(a,SZ(item[1]))];
                lans+=pre[2][min(SZ(item[2]),b+rc+rd*2)];
                ans=max(lans,ans);
            }
        }
    }
    cout<<ans<<"\n";
    // ll suma=0,idxa=a;
    // FOR(i,0,SZ(item[1])-1){
    //     suma+=item[1][i];
    // }
    
}
/*
AC
https://acm.hdu.edu.cn/contest/view-code?cid=1176&rid=11903
*/

心路历程（WA，TLE，MLE……）

2025牛客暑期多校训练营6——Problem L. 最小括号串（括号匹配，）

Posted on 2025-07-31 Edited on 2026-04-06

思路讲解

离AC之差一步之遥。

那为什么要倒着遍历那？说实话，我也不知道。

这个是

证明：上述过程相当于，一开始将所有的左括号放在前 n 个位置，即字
典序最小的合法括号串。当枚举到的区间带来硬性需求时，我们将最靠
右的左括号移动到当前区间的左端点。这样的操作显然是最优的。时间
复杂度 O(n + m log m)。

AC代码

https://ac.nowcoder.com/acm/contest/view-submission?submissionId=78588028

源代码

// by Gospel_rock
// from template_cpp
#include <bits/stdc++.h>
#define all(vec) vec.begin(),vec.end()
#define pb push_back
#define EB emplace_back
#define EM emplace
#define FI first
#define SE second
#define pct __builtin_popcountll
#define ctz __builtin_ctzll
#define SZ(a) ((int) a.size())
#define FOR(i, a, b) for (long long i = (a); i <= (b); ++i)
#define ROF(i, a, b) for (long long i = (a); i >= (b); --i)
#define debug(var) cerr << #var <<":"<<var<<"\n";
#define cend cerr<<"\n-----------\n"
#define lson(var) (var<<1)
#define rson(var) ((var<<1)+1)
#define fsp(x) fixed<<setprecision(x)

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 vdb = vector<DB>;
using tll = tuple<ll,ll,ll>;
using vpll = vector<pll>;using vtll = vector<tll>;
using arr3 = array<ll,3> ;using arr2 = array<ll,2>;
constexpr ll MAXN=(ll)1e6+10,INF=(ll)1e17+9; // 1e18+9也是素数
constexpr ll mod=(ll)1e9+7; // 998244353;
constexpr double eps=1e-8;const double pi=acos(-1.0);
ll N,M,Q,X,K,T,lT;

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

*/
// char S[MAXN];
inline void __(){
	cin>>N>>M;
    vpll lr;
    vector<char> S(2*N+5);
    vll A(2*N+5);
    bool isN=false;
	FOR(i,0,M-1){
        ll l,r;
        cin>>l>>r;
        if(l!=1){
            lr.EB(l,r);
        }
        if(l==2*N){
            isN=true;
        }
    }
    if(isN){
        cout<<-1<<"\n";
        return;
    }
    sort(all(lr),greater<>());
    vpll llr;
    if(!lr.empty()){
        ll l=lr[0].FI,r=lr[0].SE;
        FOR(i,1,SZ(lr)-1){
            ll ll=max(l,lr[i].FI),rr=min(r,lr[i].SE);
            if(ll>rr){  
                llr.EB(l,r);
                l=lr[i].FI;r=lr[i].SE;
            }else{
                l=ll;
                r=rr;
            }
        }
        llr.EB(l,r);
    }
    S[1]='(';A[1]=1;
    FOR(i,0,SZ(llr)-1){
        S[llr[i].FI]='(';
        A[llr[i].FI]=1;
    }
    vll Sum(2*N+5);
    FOR(i,1,2*N){
        if(A[i]==0){
            A[i]=-1;
        }
        Sum[i]=Sum[i-1]+A[i];
    }
    ll N2=2*N;
#ifdef LOCAL
    FOR(i,1,N2){
        cerr<<Sum[i]<<" ";
    }
    cerr<<"\n";
    // FOR(i,1,N2){
    //     cerr<<S[i]<<" ";
    // }
    // cerr<<"\n";
    FOR(i,0,SZ(llr)-1){
        debug(llr[i].FI);
        debug(llr[i].SE);
    }
#endif
    if(Sum[2*N]>0){
        cout<<-1<<"\n";
        return;
    }
    ll sum=Sum[N2];
    FOR(i,1,N2){
        if(S[i]!='('){
            if(sum<0){
                sum+=2;     // -1变成+1，+2。
                S[i]='(';
            }else{
                S[i]=')';
            }
        }
    }
    // ())((((((())))))  
    sum=0;
    FOR(i,1,N2){
        if(S[i]=='('){
            sum+=1;
        }else{
            sum-=1;
        }
        // 匹配失败（对于这个右括号，前面已经没有左括号匹配了）
        if(sum<0){
            cout<<-1<<"\n";
            return;
        }
    }
    FOR(i,1,N2){
        cout<<S[i];
    }
    cout<<"\n";
}
/*
AC 
https://ac.nowcoder.com/acm/contest/view-submission?submissionId=78588028
1
2 3
1 2
2 3
3 4

()()
*/

还有一种树状数组解法。

源代码

// by Gospel_rock
#include <bits/stdc++.h>
#define all(vec) vec.begin(),vec.end()
#define pb push_back
#define EB emplace_back
#define EM emplace
#define FI first
#define SE second
#define pct __builtin_popcountll
#define ctz __builtin_ctzll
#define SZ(a) ((int) a.size())
#define FOR(i, a, b) for (long long i = (a); i <= (b); ++i)
#define ROF(i, a, b) for (long long i = (a); i >= (b); --i)
#define debug(var) cerr << #var <<":"<<var<<"\n";
#define cend cerr<<"\n-----------\n"
#define lson(var) (var<<1)
#define rson(var) ((var<<1)+1)
#define fsp(x) fixed<<setprecision(x)

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 tll = tuple<ll,ll,ll>;
using vpll = vector<pll>;using vtll = vector<tll>;
using vdb = vector<DB>;
using arr3 = array<ll,3> ;using arr2 = array<ll,2>;
constexpr ll MAXN=(ll)4e5+10,INF=(ll)1e17+9; // 1e18+9也是素数
constexpr ll mod=(ll)1e9+7; // 998244353;
constexpr double eps=1e-8;const double pi=acos(-1.0);
ll N,M,Q,X,K,T,lT;

inline void __();
ll tr[MAXN];
ll lowbit(ll x){
    return x&(-x);
}
void add(ll p,ll x){
    for(int i=p;i<=2*N;i+=lowbit(i)){
        tr[i]+=x;
    }
}
ll query(ll p){
    ll res=0;
    for(int i=p;i>0;i-=lowbit(i)){
        res+=tr[i];
    }
    return res;
}
ll query(ll l,ll r){
    l-=1;
    ll res=query(r)-query(l);
    return res;
}
signed main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);cout.tie(0);
    
    cin>>lT;
    while(lT--)
        __();
    return 0;
}
/*

*/
inline void __(){
    cin>>N>>M;
    FOR(i,0,2*N+5){
        tr[i]=0;
    }
    vll A(2*N+5);
    vpll lr;
    FOR(i,1,M){
        ll l,r;
        cin>>l>>r;
        lr.EB(l,r);
    }
    sort(all(lr),greater<>());
    FOR(i,0,M-1){
        auto &[l,r]=lr[i];
        if(query(l,r)>0) continue;
        add(l,1);
        A[l]=1;
    }
    A[1]=1;
    ll N2=2*N;
    vll Sum(N2+5);
    FOR(i,1,N2){
        if(A[i]!=1) A[i]=-1;
        Sum[i]=Sum[i-1]+A[i];
    }
    if(Sum[N2]>0){
        cout<<-1<<"\n";
        return;
    }
    ll sum=Sum[N2];
#ifdef LOCAL
    debug(sum);
    debug(query(1,2*N));
    FOR(i,1,N2){
        cerr<<Sum[i]<<" ";
    }
    cend;
#endif
    FOR(i,1,N2){
        if(A[i]==1) continue;
        if(sum<0) sum+=2,A[i]=1;
        else A[i]=-1;
    }
    sum=0;
    FOR(i,1,N2){
        sum+=A[i];
        if(sum<0){
            cout<<-1<<"\n";
            return;
        }
    }
    FOR(i,1,N2){
        if(A[i]==1){
            cout<<"(";
        }else{
            cout<<")";
        }
    }
    cout<<"\n";
}
/*

*/