제출 #580287

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
580287		tranxuanbach	Linear Garden (IOI08_linear_garden)	C++17	41 / 100	21 ms	14108 KiB

linear_garden

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace __gnu_pbds;

#define endl '\n'
#define fi first
#define se second
#define For(i, l, r) for (auto i = (l); i < (r); i++)
#define ForE(i, l, r) for (auto i = (l); i <= (r); i++)
#define FordE(i, l, r) for (auto i = (l); i >= (r); i--)
#define Fora(v, a) for (auto v: (a))
#define bend(a) (a).begin(), (a).end()
#define isz(a) ((signed)(a).size())

using ll = long long;
using ld = long double;
using pii = pair <int, int>;
using vi = vector <int>;
using vpii = vector <pii>;
using vvi = vector <vi>;

const int N = 1e6 + 5;
int mod;

inline void sadd(int &x, int y){
    if ((x += y) >= mod){
        x -= mod;
    }
}

inline int add(int x, int y){
    if ((x += y) >= mod){
        x -= mod;
    }
    return x;
}

inline void ssub(int &x, int y){
    if ((x -= y) < 0){
        x += mod;
    }
}

inline int sub(int x, int y){
    if ((x -= y) < 0){
        x += mod;
    }
    return x;
}

int n;
string s;

int dp[3][N];

signed main(){
    ios_base::sync_with_stdio(0);
    cin.tie(0); cout.tie(0);
    // freopen("KEK.inp", "r", stdin);
    // freopen("KEK.out", "w", stdout);
    cin >> n >> mod;

    dp[-1 + 1][0] = 1;
    dp[0 + 1][0] = 1;
    dp[1 + 1][0] = 1;
    ForE(i, 1, n){
        ForE(d, -1, 1){
            if (d >= 0){
                sadd(dp[d + 1][i], dp[d - 1 + 1][i - 1]);
            }
            if (d <= 0){
                sadd(dp[d + 1][i], dp[d + 1 + 1][i - 1]);
            }
        }
    }

    cin >> s;
    int d = 0, mxd = 0, mnd = 0, ans = 1;
    For(i, 0, n){
        if (s[i] == 'P'){
            d--;
            int tmxd = max(mxd, d), tmnd = min(mnd, d);
            if (-2 <= tmnd and tmxd <= 0){
                sadd(ans, dp[(d + 1) + 1][n - i - 1]);
            }
            if (-1 <= tmnd and tmxd <= 1){
                sadd(ans, dp[d + 1][n - i - 1]);
            }
            if (0 <= tmnd and tmxd <= 2){
                sadd(ans, dp[(d - 1) + 1][n - i - 1]);
            }
            if (-1 <= tmnd and tmxd <= 0){
                ssub(ans, 1);
            }
            if (0 <= tmnd and tmnd <= 1){
                ssub(ans, 1);
            }
            d++;
        }
        if (s[i] == 'L'){
            d--;
        }
        else{
            d++;
        }
        mxd = max(mxd, d);
        mnd = min(mnd, d);
    }
    cout << ans << endl;
}

/*
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INPUT:                                            |
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OUTPUT:                                           |
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*/

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