結果

問題 No.3510 RPS Eliminations
コンテスト
ユーザー ponjuice
提出日時 2026-04-18 23:29:56
言語 C++23
(gcc 15.2.0 + boost 1.89.0)
コンパイル:
g++-15 -O2 -lm -std=c++23 -Wuninitialized -DONLINE_JUDGE -o a.out _filename_
実行:
./a.out
結果
AC  
実行時間 99 ms / 2,000 ms
コード長 7,377 bytes
記録
記録タグの例:
初AC ショートコード 純ショートコード 純主流ショートコード 最速実行時間
コンパイル時間 4,037 ms
コンパイル使用メモリ 365,092 KB
実行使用メモリ 51,764 KB
最終ジャッジ日時 2026-04-18 23:30:13
合計ジャッジ時間 11,752 ms
ジャッジサーバーID
（参考情報） 		judge1_0 / judge3_1
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 1
other AC * 28
権限があれば一括ダウンロードができます

ソースコード

diff #
raw source code 

#include <bits/stdc++.h>
using namespace std;

//高速化 
struct ponjuice{ponjuice(){cin.tie(0);ios::sync_with_stdio(0);cout<<fixed<<setprecision(20);}}PonJuice;
//#define endl '\n' //インタラクティブ問題の時は消す

//型
using ll = long long;
using ld = long double;
template<class T>using vc = vector<T>; template<class T>using vvc = vc<vc<T>>; template<class T>using vvvc = vvc<vc<T>>;
using vi = vc<int>;  using vvi = vvc<int>;  using vvvi = vvvc<int>;
using vl = vc<ll>;   using vvl = vvc<ll>;   using vvvl = vvvc<ll>;
using pi = pair<int, int>;  using pl = pair<ll, ll>;
using ull = unsigned ll;
template<class T>using priq = priority_queue<T>;
template<class T>using priqg = priority_queue<T, vc<T>, greater<T>>;

// for文
#define overload4(a, b, c, d, e, ...) e
#define rep1(n)             for(ll i = 0; i < n; i++)
#define rep2(i, n)          for(ll i = 0; i < n; i++)
#define rep3(i, a, b)       for(ll i = a; i < b; i++)
#define rep4(i, a, b, step) for(ll i = a; i < b; i+= step)
#define rep(...) overload4(__VA_ARGS__, rep4, rep3, rep2, rep1)(__VA_ARGS__)
#define per1(n)             for(ll i = n-1; i >= 0; i--)
#define per2(i, n)          for(ll i = n-1; i >= 0; i--)
#define per3(i, a, b)       for(ll i = b-1; i >= a; i--)
#define per4(i, a, b, step) for(ll i = b-1; i >= a; i-= step)
#define per(...) overload4(__VA_ARGS__, per4, per3, per2, per1)(__VA_ARGS__)

//関数
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
#define si(x) (ll)(x).size()
template<class S, class T>inline bool chmax(S& a, T b){return a < b && ( a = b , true);}
template<class S, class T>inline bool chmin(S& a, T b){return a > b && ( a = b , true);}

inline void yes(){cout << "Yes\n";}
inline void no(){cout << "No\n";}
inline void yesno(bool y = true){if(y)yes();else no();}

//定数
constexpr ll mod = 998244353;
constexpr ll minf=-(1<<29);
constexpr ll inf=(1<<29);
constexpr ll MINF=-(1LL<<60);
constexpr ll INF=(1LL<<60);
const int dx[4] ={-1, 0, 1, 0};
const int dy[4] ={ 0, 1, 0,-1};
const int dx8[8] ={-1,-1,-1, 0, 1, 1, 1, 0};
const int dy8[8] ={-1, 0, 1, 1, 1, 0,-1,-1};

string solve(const ll n,const ll k, vector<ll> a);

ll powll(ll x, ll n) {
    ll res = 1;
    while(n > 0) {
        if(n & 1) res *= x;
        x *= x;
        n >>= 1;
    }
    return res;
}

int battle(string s, const vector<ll>& a) {
    map<char,int> mp;
    mp['P'] = 0;
    mp['R'] = 1;
    mp['S'] = 2;
    vector<int> now;
    rep(i,0,s.size()) now.push_back(mp[s[i]]);
    int n = now.size();

    rep(i,0,n-1) {
        vector<int> nx;
        rep(j,0,now.size()) {
            if(a[i]-1 == j) {
                if(now[j] == now[j+1]) return -1;
                if((now[j]+1)%3 == now[j+1]) nx.push_back(now[j]);
                else nx.push_back(now[j+1]);
                j++;
            }
            else {
                nx.push_back(now[j]);
            }
        }
        now.swap(nx);
    }
    return now[0];
}

string greedy(ll n, ll k, vector<ll> a) {
    string s(n, '?');
    vector<vector<string>> ans(3);
    rep(i,0,powll(3,n)) {
        int I = i;
        rep(j,0,n) {
            s[j] = "PRS"[I%3];
            I /= 3;
        }
        int win = battle(s, a);
        if(win != -1) ans[win].push_back(s);
    }
    if(k >= ans[0].size()) {
        return "-1\n-1\n-1\n";
    }
    sort(all(ans[0]));
    sort(all(ans[1]));
    sort(all(ans[2]));
    return ans[1][k] + "\n" + ans[0][k] + "\n" + ans[2][k] + "\n"; 
}

void test() {
    int itr = 0;
    while(true){
        cout << (itr++) << endl;
        ll n = 10, k = rand() % 16;
        vector<ll> a(n-1);
        rep(i,0,n-1) a[i] = rand() % (n-i-1) + 1;
        if(greedy(n,k,a) != solve(n,k,a)) {
            cout << n << " " << k << endl;
            for(auto c: a) cout << c << " "; cout << endl;

            cout << greedy(n,k,a) + "\n" + solve(n,k,a) << endl;
            return;
        } 

        //     cout << greedy(n,k,a) + "\n" + solve(n,k,a) << endl;
        // break;
    }
}

int main() {
    // test();
    // return 0;

	int t = 1;
    cin >>t;
    while(t--){
        ll n,k;
        cin >> n >> k;k--;
        vector<ll> a(n-1);
        rep(i,0,n-1) cin >> a[i];
        cout << solve(n,k,a);
    }
}

/*
せっかくなので、PPC は全てコメント書きます



辞書順でK番目系のやつ苦手

木DP 的な感じかな？
とりあえず木DPだと思って解いてみます

REだけになってくれて一安心
*/

#include<atcoder/segtree>
using namespace atcoder;

int add(int a, int b) {return a + b;}
int zero() {return 0;}

ll sani(ll a, ll b) {
    if(INF/a <= b) return INF;
    return a * b;
}

string solve(const ll n,const ll k, vector<ll> a){

    // 木の構築
    segtree<int,add,zero> seg(n);
    rep(i,0,n) seg.set(i,1);

    vector<int> node(n);
    
    rep(i,0,n) node[i] = i;
    
    vector<ll> l(n*2, -1);
    vector<ll> r(n*2, -1);
    vector<ll> cnt(n*2, 1);

    rep(i,0,n-1) {
        int L = seg.max_right(0, [&](int x){return x < a[i];});
        int R = seg.max_right(0, [&](int x){return x < a[i]+1;});
        seg.set(R, 0);
        l[n+i] = node[L];
        r[n+i] = node[R];
        cnt[n+i] = cnt[node[L]] + cnt[node[R]];
        chmin(cnt[n+i], 61);
        node[L] = n+i;
    }


    using S = array<ll,3>; // P R S
    auto sol = [&](S st) -> string {
        // この中にありそう
        string ans(n, '?');
        bool possible = true;

        auto dfs = [&](auto&&self, int nw, S p, ll K) -> pair<int,ll> {
            // cout << nw << " " << K << "  p:[" << p[0] << "," << p[1] << "," << p[2] << "]" << endl;
            if(nw < n) {
                if(K < p[0]) {
                    ans[nw] = 'P';
                    return {0, 0};
                }
                K -= p[0];
                if(K < p[1]) {
                    ans[nw] = 'R';
                    return {1, p[0]};
                }
                K -= p[1];
                if(K < p[2]) {
                    ans[nw] = 'S';
                    return {2, p[0]+p[1]};
                }
                possible = false;
                return {0, 0};
            }
            ll lk = K / (1LL << min(60LL, cnt[r[nw]]-1));
            S lp = p;
            rep(i,0,3) {
                lp[(i+1)% 3] += p[i];
                chmin(lp[(i+1)% 3], k+1);
            }

            auto [lc, mc] = self(self, l[nw], lp, lk);
            if(!possible) return {0,0};
            mc *= (1LL << min(60LL, cnt[r[nw]]-1));
            // cout << nw << " " << "PRS"[lc] << " " << mc << endl;
            K -= mc;

            S rp = {0, 0, 0};
            rp[(lc+1)%3] = p[lc%3];
            rp[(lc+2)%3] = p[(lc+2)%3];

            ll rk = K % sani((1LL << min(60LL, cnt[r[nw]]-1)), lp[lc]); // これ % じゃないかも

            auto [rc, mc2]= self(self, r[nw], rp, rk);
            if(!possible) return {0,0};
            // cout << nw << " " << "PRS"[rc] << " " << mc << endl;
            mc += mc2;

            if((lc+1)%3 == rc) return {lc, mc};
            return {rc, mc};
        };

        dfs(dfs, n*2-2, st, k);
        // cout << endl;
        if(possible) return ans;
        else return "-1";
    };

    string ans1 = sol({0, 1, 0});
    string ans2 = sol({1, 0, 0});
    string ans3 = sol({0, 0, 1});

    return ans1 + "\n" + ans2 + "\n" + ans3 + "\n"; 
}

/*

RPRSP
PPSSR
PPRSP

*/
0