Submission #1154164

#	Time	Username	Problem	Language	Result	Execution time	Memory
1154164		pan	Binaria (CCO23_day1problem1)	C++20	25 / 25	198 ms	31732 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 d[1000005], root[1000005], ans[1000005], fact[1000005];

ll mod = 1e6 + 3;
// inverse modulo by extended euclidean's algorithm
 
ll inv(ll a) {
    return a <= 1 ? a : mod - (ll)(mod / a) * inv(mod % a) % mod;
}

ll ncr(ll n, ll r)
{
	return fact[n]*inv(fact[n-r])%mod*inv(fact[r])%mod;
}


int main()
{
	fact[0] = 1;
	for (ll i=1; i<=1000000; ++i) fact[i] = fact[i-1]*i%mod;
	ll n, k;
	cin >> n >> k;
	for (ll i=0; i<n; ++i) ans[i] = -1;
	for (ll i=0; i<k; ++i) root[i] = i;
	for (ll i=0; i<n-k+1; ++i) cin >> d[i];
	for (ll i=k; i<n; ++i)
	{
		ll pos = i-k + 1;
		if (d[pos] - d[pos-1] == 0) root[i] = root[i-k];
		else if (d[pos] - d[pos-1] == 1) 
		{
			if (ans[root[i-k]] == 1) {cout << "0\n"; return 0;}
			ans[root[i-k]] = 0;
			root[i] = i;
			ans[i] = 1;
		}
		else if (d[pos] - d[pos-1] == -1)
		{
			if (ans[root[i-k]] == 0) {cout << "0\n"; return 0;}
			ans[root[i-k]] =1;;
			root[i] = i;
			ans[i] = 0;
		}
		else {cout << "0\n"; return 0;}
	}
	//for (ll i=0; i<k; ++i) show(ans[i]);
	ll cc = 0, nn = d[0];
	for (ll i=0; i<k; ++i)
	{
		if (ans[i]==-1) cc++;
		else if (ans[i]==1) nn--;
	}
	//show2(nn, cc);
	cout << ncr(cc, nn) << endl;
	return 0;
}

• Subtask 1: 3 / 3
• Subtask 2: 3 / 3
• Subtask 3: 4 / 4
• Subtask 4: 4 / 4
• Subtask 5: 4 / 4
• Subtask 6: 7 / 7