Submission #38185

#	Time	Username	Problem	Language	Result	Execution time	Memory
38185		farmersrice	Calvinball championship (CEOI15_teams)	C++14	100 / 100	199 ms	2488 KiB

teams

#include <bits/stdc++.h>
//#pragma GCC optimize ("O3")
//#pragma GCC target ("sse4")
#pragma GCC target ("avx,tune=native")
//Use above if bruteforcing with lots of small operations. Or just use it anytime, there's no downside. AVX is better slightly
/*
TASK: hidden
LANG: C++11
*/
using namespace std;
typedef long long ll;
typedef pair<int, int> pair2;
typedef pair<int, pair<int, int> > pair3;
typedef pair<int, pair<int, pair<int, int> > > pair4;
#define MAXN 10013
#define INF 1000000000000000000LL
#define mp make_pair
#define add push_back
#define remove pop
#define MOD 1000007LL

ll n;
ll values[MAXN];
ll num[MAXN];
ll dp[2][MAXN]; //dp[i][j]: num ways from point i IF the highest number that appears before us is j (implying value of i is at most j + 1)
//dp[i][j] = dp[i + 1][j] * j + dp[i + 1][j + 1]

int main() {
	ios_base::sync_with_stdio(false); 
	cin.tie(NULL);

	cin >> n;

	for (int i = 0; i < n; i++) {
		cin >> values[i];
	}	

	for (int i = 0; i < 2; i++) {
		for (int j = 0; j < MAXN; j++) {
			dp[i][j] = 1;
		}
	}

	int answer = 0;
	ll highestNum = 0;

	for (int i = 0; i < n; i++) {
		num[i] = highestNum;
		highestNum = max(highestNum, values[i]);
	}

	int old = 0;
	for (int i = n - 1; i >= 0; i--) {
		answer += ((values[i] - 1) * dp[old][num[i]]) % MOD;
		answer %= MOD;

		for (int j = 0; j < i; j++) {
			dp[1 - old][j] = dp[old][j] * j + dp[old][j + 1];
			dp[1 - old][j] %= MOD;
		}
		old = 1 - old;
	}

	cout << answer + 1 << endl;
}

• Subtask 1: 10 / 10
• Subtask 2: 10 / 10
• Subtask 3: 10 / 10
• Subtask 4: 10 / 10
• Subtask 5: 10 / 10
• Subtask 6: 10 / 10
• Subtask 7: 10 / 10
• Subtask 8: 10 / 10
• Subtask 9: 10 / 10
• Subtask 10: 10 / 10