#1160107 (C++23) No.3510 RPS Eliminations

• 問題: No.3510 RPS Eliminations
• ユーザー: ponjuice
• 提出日時: 2026-04-18 21:59:16
• 言語: C++23 (gcc 15.2.0 + boost 1.89.0)
• 結果: RE
• 実行時間: -
• コード長: 4,914 bytes
• コンパイル時間: 2,702 ms
• コンパイル使用メモリ: 341,648 KB
• 実行使用メモリ: 48,660 KB
• 最終ジャッジ日時: 2026-04-18 21:59:34
• 合計ジャッジ時間: 7,187 ms
• ジャッジサーバーID （参考情報）: judge2_0 / judge3_1

ファイルパターン	結果
sample	AC * 1
other	WA * 23 RE * 5

ソースコード

#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};

void solve();
int main() {
	int t = 1;
    cin >>t;
    while(t--){
        solve();
    }
}

/*
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
*/

#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;
}

void solve(){
    ll n,k;
    cin >> n >> k;
    k--;
    vector<ll> a(n-1);
    rep(i,0,n-1) cin >> a[i];

    // 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 ans(n, '?');
        bool possible = true;

        auto dfs = [&](auto&&self, int nw, S p, ll K) -> int {
            // cout << nw << " " << K << "  p:[" << p[0] << "," << p[1] << "," << p[2] << "]" << endl;
            if(nw < n) {
                if(K < p[0]) {
                    ans[nw] = 'P';
                    return 0;
                }
                K -= p[0];
                if(K < p[1]) {
                    ans[nw] = 'R';
                    return 1;
                }
                K -= p[1];
                if(K < p[2]) {
                    ans[nw] = 'S';
                    return 2;
                }
                possible = false;
                return 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);
            }

            int lc = self(self, l[nw], lp, lk);
            // cout << nw << " " << "PRS"[lc] << endl;

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

            ll rk = K % sani((1LL << min(60LL, cnt[r[nw]]-1)), lp[lc]);

            int rc = self(self, r[nw], rp, rk);
            // cout << nw << " " << "PRS"[rc] << endl;

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

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

    sol({0, 1, 0});
    sol({1, 0, 0});
    sol({0, 0, 1});
    // cout << endl;
}