MX 2025--炼石计划 NOIP 模拟赛 #20

打得抽象。T3,T4放俩难的板子。由于是MX的题，就不放题意了。

邻间的骰子之舞

发现复制操作不会超过\(64\)次，而粘贴操作肯定是越均匀越好，直接二分暴力跑就行了。

点此查看代码

#include<bits/stdc++.h>
using namespace std;
#define rep(i,s,t,p) for(int i = s;i <= t; i += p)
#define drep(i,s,t,p) for(int i = s;i >= t; i -= p)
#ifdef LOCAL
    FILE *InFile = freopen("in.in","r",stdin),*OutFile = freopen("out.out","w",stdout);
#else
    // FILE *InFile = stdin,*OutFile = stdout;
    FILE *InFile = freopen("dice.in","r",stdin),*OutFile = freopen("dice.out","w",stdout);
#endif
using ll=long long;using ull=unsigned long long;
using db = double;using ldb = long double;
#define int ull
int n,x,y,mx;
inline bool check(int mid){
    rep(i,0,mx,1){
        if(i*x >= mid) break;
        if(!i){
            if((mid - i*x)/y + 1 > n) return true;
            continue;
        }
        int j = (mid - i * x) / y,val = 1,aver = j / (i + 1),last = j-aver*(i + 1);
		rep(k,0,i,1){
			if (last > 0) val += (aver + 1)*val,last --;
            else val += aver * val;
			if (val > n) return true; 
		}
    }
    return false;
}
inline void solve(){
    cin>>n>>x>>y;mx = log2(n);
    int l = 1,r = 1e19,ans = 0;
    while(l <= r){
        int mid = (l + r) >> 1;
        if(check(mid)) ans = mid,r = mid - 1;
        else l = mid + 1;
    }
    cout<<ans+x<<'\n';
}
signed main(){
    cin.tie(nullptr)->sync_with_stdio(false);
    solve();
}

星海浮沉录

用\(set\)维护每个数出现的位置，用\(multiset\)维护每个数可以贡献的长度，用线段树维护区间最大值，线段树上二分即可。\(set\)插入哨兵节点会更方便一点。

点此查看代码

#include<bits/stdc++.h>
using namespace std;
#define rep(i,s,t,p) for(int i = s;i <= t; i += p)
#define drep(i,s,t,p) for(int i = s;i >= t; i -= p)
#ifdef LOCAL
    FILE *InFile = freopen("in.in","r",stdin),*OutFile = freopen("out.out","w",stdout);
#else
    // FILE *InFile = stdin,*OutFile = stdout;
    FILE *InFile = freopen("star.in","r",stdin),*OutFile = freopen("star.out","w",stdout);
#endif
using ll=long long;using ull=unsigned long long;
using db = double;using ldb = long double;
const int N = 5e5 + 10;
int n,q,a[N];
set<int> pos[N];
multiset<int> mx[N];
#define ins insert
struct Segment_Tree{
    struct segment_tree{
        int l,r,val;
        #define l(x) tree[x].l
        #define r(x) tree[x].r
        #define val(x) tree[x].val
    }tree[N<<2];
    inline void pushup(int k){val(k) = max(val(k<<1),val(k<<1|1));}
    void build(int k,int l,int r){
        l(k) = l,r(k) = r;
        if(l == r) return val(k) = *mx[l].rbegin(),void();
        int mid = (l + r) >> 1;
        build(k<<1,l,mid);build(k<<1|1,mid+1,r);
        pushup(k);
    }
    void upd(int k,int pos,int val){
        if(l(k) == r(k)) return val(k) = val,void();
        int mid = (l(k) + r(k)) >> 1;
        if(pos <= mid) upd(k<<1,pos,val);else upd(k<<1|1,pos,val);
        pushup(k);
    }
    int qry(int k,int x){
        if(l(k) == r(k)) return l(k);
        int ls = k<<1;
        if(val(ls) >= x) return qry(k<<1,x);
        else return qry(k<<1|1,x);
    }
}T;
inline void solve(){
    cin>>n>>q;
    rep(i,1,n,1) cin>>a[i];
    rep(i,0,n+1,1) pos[i].ins(0);
    rep(i,1,n,1){
        auto it = pos[a[i]].ins(i).first;
        it--;mx[a[i]].ins(i - *it - 1);
    }
    rep(i,0,n+1,1){
        auto it = pos[i].ins(n+1).first;
        it--;mx[i].ins(n - *it);
    }
    T.build(1,0,n+1);
    auto del = [&](int val,int s){
        auto it = pos[val].lower_bound(s);
        int p1,p2;
        it--;mx[val].erase(mx[val].lower_bound(s-*it-1));p1 = *it;it++;
        it++;mx[val].erase(mx[val].lower_bound(*it-s-1));p2 = *it;
        mx[val].insert(p2-p1-1);
        pos[val].erase(s);
    };
    auto add = [&](int val,int s){
        auto it = pos[val].upper_bound(s);
        int p1,p2;
        p2 = *it;it--;p1 = *it;
        mx[val].erase(mx[val].lower_bound(p2-p1-1));
        mx[val].insert(p2-s-1);
        mx[val].insert(s-p1-1);
        pos[val].ins(s);
    };
    auto change = [&](int val,int s,int t){
        del(val,s);add(val,t);
        T.upd(1,val,*mx[val].rbegin());
    };
    rep(test,1,q,1){
        int op,x;cin>>op>>x;
        if(op == 1){
            if(a[x] == a[x+1]) continue;
            change(a[x],x,x+1);
            change(a[x+1],x+1,x);
            swap(a[x],a[x+1]);
        }
        else cout<<T.qry(1,x)<<'\n';
    }
}
signed main(){
    cin.tie(nullptr)->sync_with_stdio(false);
    solve();
}

勾指起誓

如题目名，勾誓。FMT板子，不会。

第八交响曲

双调排序板子，不会的自行bdfs吧。

点此查看代码

#include<bits/stdc++.h>
using namespace std;
#define rep(i,s,t,p) for(int i = s;i <= t; i += p)
#define drep(i,s,t,p) for(int i = s;i >= t; i -= p)
#ifdef LOCAL
    FILE *InFile = freopen("in.in","r",stdin),*OutFile = freopen("out.out","w",stdout);
#else
    // FILE *InFile = stdin,*OutFile = stdout;
    FILE *InFile = freopen("symphony.in","r",stdin),*OutFile = freopen("symphony.out","w",stdout);
#endif
using ll=long long;using ull=unsigned long long;
using db = double;using ldb = long double;
int n,lgn;
inline void solve(){
    cin>>n;lgn = 1<<__lg(n-1)+1;
    auto print = [](int a,int b){if(b <= n) cout<<"CMPSWP R"<<a<<" R"<<b<<' ';};
    cout<<__lg(lgn)*(__lg(lgn)+1)/2<<'\n';
    rep(i,1,lgn-1,i){
        rep(j,1,lgn,i*2) rep(k,0,i-1,1) print(j+k,j+i*2-1-k);
        cout<<'\n';
        for(int j = i/2;j >= 1;j /= 2){
            rep(k,1,lgn,j*2) rep(p,0,j-1,1) print(k+p,k+j+p);
            cout<<'\n';
        }
    }
}
signed main(){
    cin.tie(nullptr)->sync_with_stdio(false);
    solve();
}

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