Submission #1181249

#	Time	Username	Problem	Language	Result	Execution time	Memory
1181249		TudorMa	Bitaro the Brave (JOI19_ho_t1)	C++20	100 / 100	266 ms	79840 KiB

joi2019_ho_t1

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

#define fi first
#define sc second
#define pb push_back

using namespace std;
using namespace __gnu_pbds;

typedef long long ll;
typedef double db;
typedef pair<int, int> pii;
template<typename type>
using ordered_set = tree<type, null_type, less<type>, rb_tree_tag, tree_order_statistics_node_update>;

const int N = 3005, mod = 1e9 + 7, inf = 2e9;
const int dl[] = {-1, 0, 1, 0}, dc[] = {0, 1, 0, -1};
const int ddl[] = {-1, -1, -1, 0, 1, 1, 1, 0}, ddc[] = {-1, 0, 1, 1, 1, 0, -1, -1};

mt19937 gen(chrono::steady_clock::now().time_since_epoch().count());
int rng(int lo = 1, int hi = INT_MAX) {
    uniform_int_distribution<int> rnd(lo, hi);
    return rnd(gen);
}

struct mint {
    int val;
    mint(int32_t x = 0) {
        val = x % mod;
    }
    mint(long long x) {
        val = x % mod;
    }
    mint operator+(mint x) {
        return val + x.val;
    }
    mint operator-(mint x) {
        return val - x.val + mod;
    }
    mint operator*(mint x) {
        return 1LL * val * x.val;
    }
    void operator+=(mint x) {
        val = (*this + x).val;
    }
    void operator-=(mint x) {
        val = (*this - x).val;
    }
    void operator*=(mint x) {
        val = (*this * x).val;
    }
    friend auto operator>>(istream& in, mint &x) -> istream& {
        in >> x.val;
        x.val %= mod;
        return in;
    }
    friend auto operator<<(ostream& out, mint const &x) -> ostream& {
        out << x.val;
        return out;
    }
};

int n, m, ln[N][N], col[N][N];
char a[N][N];

int32_t main()
{
    cin.tie(nullptr)->sync_with_stdio(0);
    
    cin >> n >> m;
    for(int i=1; i<=n; i++)
        for(int j=1; j<=m; j++)
            cin >> a[i][j];
    
    for(int i=1; i<=n; i++)
        for(int j=1; j<=m; j++)
            ln[i][j] = ln[i][j-1] + (a[i][j] == 'O');
    for(int j=1; j<=m; j++)
        for(int i=1; i<=n; i++)
            col[i][j] = col[i-1][j] + (a[i][j] == 'I');

    ll ans = 0;
    for(int i=1; i<=n; i++)
        for(int j=1; j<=m; j++)
            if(a[i][j] == 'J')
                ans += 1LL * (ln[i][m] - ln[i][j]) * (col[n][j] - col[i][j]);
    cout << ans;
    return 0;
}

• Subtask 1: 20 / 20
• Subtask 2: 30 / 30
• Subtask 3: 50 / 50