Submission #1275370

#	Time	Username	Problem	Language	Result	Execution time	Memory
1275370		cngkhanh_ti22	Linear Garden (IOI08_linear_garden)	C++20	100 / 100	45 ms	5340 KiB

linear_garden

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

// Flattened index for state (hl in [0..2], hp in [0..2], ok in [0..1])
inline int idx(int hl, int hp, int ok) {
    return (hl * 3 + hp) * 2 + ok;
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    int n;
    int m;
    if(!(cin >> n >> m)) return 0;
    string s;
    cin >> s; // expected length n

    // Convert to 0 for 'L' and 1 for 'P'
    vector<int> a(n);
    for (int i = 0; i < n; ++i) a[i] = (s[i] == 'P');

    const int STATES = 3 * 3 * 2; // 18
    vector<int> next(STATES), cur(STATES);

    // Base: DP at position n+1 (i > n) is 1 for every (hl,hp,ok)
    for (int t = 0; t < STATES; ++t) next[t] = 1 % m;

    // iterate i from n down to 1 (we use zero-based index i = n-1 ... 0)
    for (int i = n - 1; i >= 0; --i) {
        int si = a[i]; // target char at position i (0 for L, 1 for P)

        // compute cur = dp[i]
        for (int hl = 0; hl <= 2; ++hl) {
            for (int hp = 0; hp <= 2; ++hp) {
                for (int ok = 0; ok <= 1; ++ok) {
                    int value = 0;

                    // Option 1: place 'L' (chosenChar = 0)
                    if (hl < 2) {
                        int okNext = ok | (0 < si); // becomes 1 if chosenChar < target
                        int nhl = hl + 1;
                        int nhp = max(0, hp - 1);
                        value = next[idx(nhl, nhp, okNext)];
                    }

                    // Option 2: place 'P' (chosenChar = 1)
                    // allowed only if (already smaller) OR (target char is 'P')
                    if (((ok == 1) || (ok == 0 && si == 1)) && (hp < 2)) {
                        int okNext = ok | (1 < si); // (1 < si) is always 0 here so okNext == ok
                        int nhl = max(0, hl - 1);
                        int nhp = hp + 1;
                        value += next[idx(nhl, nhp, okNext)];
                        if (value >= m) value -= m;
                    }

                    cur[idx(hl, hp, ok)] = value % m;
                }
            }
        }

        // roll
        next.swap(cur);
    }

    // result is dp[1][0][0][0] i.e. dp at index 0 with hl=0,hp=0,ok=0
    int result = next[idx(0,0,0)] % m;
    if (result < 0) result += m;
    cout << result << '\n';
    return 0;
}

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