제출 #1160306

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
1160306		pan	Kamenčići (COCI21_kamencici)	C++20	0 / 70	1 ms	1600 KiB

Main

#include <bits/stdc++.h>
//#include "includeall.h"
//#include <ext/pb_ds/assoc_container.hpp>
//#include <ext/pb_ds/tree_policy.hpp>
#define endl '\n'
#define f first
#define s second
#define pb push_back
#define mp make_pair
#define lb lower_bound
#define ub upper_bound
#define input(x) scanf("%lld", &x);
#define input2(x, y) scanf("%lld%lld", &x, &y);
#define input3(x, y, z) scanf("%lld%lld%lld", &x, &y, &z);
#define input4(x, y, z, a) scanf("%lld%lld%lld%lld", &x, &y, &z, &a);
#define print(x, y) printf("%lld%c", x, y);
#define show(x) cerr << #x << " is " << x << endl;
#define show2(x,y) cerr << #x << " is " << x << " " << #y << " is " << y << endl;
#define show3(x,y,z) cerr << #x << " is " << x << " " << #y << " is " << y << " " << #z << " is " << z << endl;
#define all(x) x.begin(), x.end()
#define discretize(x) sort(x.begin(), x.end()); x.erase(unique(x.begin(), x.end()), x.end());
#define FOR(i, x, n) for (ll i =x; i<=n; ++i) 
#define RFOR(i, x, n) for (ll i =x; i>=n; --i) 
#pragma GCC optimize("O3","unroll-loops")
using namespace std;
mt19937_64 rnd(chrono::steady_clock::now().time_since_epoch().count());
//using namespace __gnu_pbds;
//#define ordered_set tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update>
//#define ordered_multiset tree<int, null_type, less_equal<int>, rb_tree_tag, tree_order_statistics_node_update>
typedef long long ll;
typedef long double ld;
typedef pair<ld, ll> pd;
typedef pair<string, ll> psl;
typedef pair<ll, ll> pi;
typedef pair<pi, ll> pii;
typedef pair<pi, pi> piii;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

ll n, k;
ll ps[400], dp[400][400][400];
string ss;

ll dfs(ll l, ll r, ll c)
{
	if (dp[l][r][c]!=-1) return dp[l][r][c];
	ll a = ps[l-1] + ps[n] - ps[r] - c;
	if (a >= k) return 0;
	if (c >= k) return 1;
	bool turn = (l + (n - r -1))&1;
	ll res;
	if (!turn) // first
	{
		res = max(dfs(l + 1, r, c + (ss[l]=='C')), dfs(l, r-1, c + (ss[r]=='C')));
	}
	else  // second
	{
		res = min(dfs(l + 1, r, c), dfs(l, r-1, c));
	}
	return dp[l][r][c] = res;
}


int main()
{
	cin >> n >> k >> ss;
	ss = '0' + ss;
	for (ll i=1; i<=n; ++i) ps[i] = ps[i-1] + ll(ss[i]=='C');
	for (ll i=0; i<=n; ++i) for (ll j=0; j<=n; ++j) for (ll c=0; c<=n; ++c) dp[i][j][c] = -1;
	if (dfs(1, n, 0)) cout << "DA\n";
	else cout << "NE\n";
	return 0;
}

• Subtask 1: 0 / 10
• Subtask 2: 0 / 20
• Subtask 3: 0 / 40