Submission #1199070

#	Time	Username	Problem	Language	Result	Execution time	Memory
1199070		_noob	Linear Garden (IOI08_linear_garden)	C++20	100 / 100	13 ms	9188 KiB

linear_garden

#pragma GCC optimize("Ofast")
#include <iostream>
#include <string>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <vector>
#include <set>
#include <map>
#include <deque>
#include <stack>
#include <queue>
#include <algorithm>
#include <cassert>
#include <random>
#include <chrono>
#include <iomanip>
#include <cmath>
#include <bitset>
#include <functional>
#include <numeric>
#define int long long
#define double long double
#define ii pair<int,int>
#define iii pair<int, ii >
#define fi first
#define se second
#define getbit(x,y) (((x)>>(y))&1ll)
#define turnon(x,y) ((x)|(1ll<<y))
#define turnof(x,y) ((x)^(1ll<<y))
#define oo 1e18
#define pb push_back
#define all(x) x.begin(),x.end()
#define con(mask) mask=(mask-1)&mask
#define Unique(val) val.erase(unique(val.begin(),val.end()),val.end())
#define ll long long
#define rand_int mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
#define rand_ll mt19937_64 rnd(chrono::steady_clock::now().time_since_epoch().count());

const int mod = 1e9 + 7;
const double pi = acos(-1);
const int N = 450;
const double esp = 1e-9;
int base = 100000;

using namespace std;

int n, m;
string s;
int f[1000005];
void add(int &x, int y) {
    x += y;
    if(x >= m) x -= m;
}
void sub(int &x, int y) {
    x -= y;
    if(x < 0) x += m;
}

int dp[2][5][5][5][2];
// dp[i][sum][mx][mn][is_less]

void sol_1() {
    dp[0][2][2][2][0] = 1;
    for(int i = 0; i < n; i++) {
        int nx = 1 - i % 2;
        int cur = i % 2;
        for(int sum = 0; sum <= 4; sum++) {
            for(int mx = 2; mx <= 4; mx++) {
                for(int mn = 0; mn <= 2; mn++) {
                    for(int is_less = 0; is_less <= 1; is_less++) {
                        int new_sum = sum + 1;
                        int new_mx = max(mx, new_sum);
                        int new_mn = min(mn, new_sum);

                        if(new_mx - new_mn <= 2) {
                            // set L
                            add(dp[nx][new_sum][new_mx][new_mn][is_less || (s[i] == 'P')], dp[cur][sum][mx][mn][is_less]);
                        }

                        new_sum = sum - 1;
                        new_mx = max(mx, new_sum);
                        new_mn = min(mn, new_sum);
                        if(new_mx - new_mn <= 2 && (is_less || (s[i] == 'P'))) {
                            // set P
                            add(dp[nx][new_sum][new_mx][new_mn][is_less], dp[cur][sum][mx][mn][is_less]);
                        }
                    }
                }
            }
        }
        memset(dp[cur], 0, sizeof(dp[cur]));
    }

    int ans = 0;
    for(int sum = 0; sum <= 4; sum++) {
        for(int mx = 2; mx <= 4; mx++) {
            for(int mn = 0; mn <= 2; mn++) {
                for(int is_less = 0; is_less <= 1; is_less++) {
                    if(mx - mn <= 2) {
                        add(ans, dp[n % 2][sum][mx][mn][is_less]);
                    }
                }
            }
        }
    }

    cout << ans;
}

void sol_2() {

    f[0] = 1;
    for(int i = 1; i <= n; i++) f[i] = f[i - 1] * 2 % m;
    /*

    Có 3 TH có thể xảy ra:
    1. Nếu đoạn mình đi trong đoạn 0 -> 2, xuất phát ở 0 => Bước chẵn có 2TH, còn lẻ chỉ có 1TH duy nhất
    2. Nếu đoạn mình đi là -2 -> 0 => đối xứng với TH1
    3. Nếu đoạn mình đi là -1 -> 1 => Bước lẻ có 2TH, còn chẵn chỉ có 1TH duy nhất
    
    */

    int sum = 0, mx = 0, mn = 0;
    int res = 1;

    for(int i = 0; i < n; i++) {
        int len = n - i - 1;
        if(s[i] == 'P') {
            // set L


            int tmp = sum + 1;
            int new_mx = max(mx, tmp);
            int new_mn = mn;

            if(new_mx - new_mn == 2) {
                // Đã lấy 2 hàng => xét vị trí bắt đầu
                if(tmp == new_mx - 1) {
                    add(res, f[(len + 1) / 2]);
                }
                else {
                    add(res, f[len / 2]);
                }
            }
            else if(new_mx - new_mn == 1) {
                add(res, f[len / 2]);
                add(res, f[(len + 1) / 2]);
                sub(res, 1);
            }

            sum--;
            mn = min(mn, sum);
        } 
        else {
            sum++;
            mx = max(mx, sum);
        }
    }
    cout << res;
}

void solve() {

    cin >> n >> m >> s;

    // sol_1();

    sol_2();

}

signed main() {





    ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);

    int t = 1;
    //cin >> t;


    while(t--) {
        solve();
    }
}
//      ProTeam
//(¯`·.·´¯) (¯`·.·´¯)
//`·.¸(¯`·.·´¯)¸ .·
//×°× ` ·.¸.·´ ×°×

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