Submission #39828

#	Time	Username	Problem	Language	Result	Execution time	Memory
39828		14kg	Boat (APIO16_boat)	C++11	9 / 100	8 ms	13092 KiB

boat

#include <stdio.h>
#include <set>
#include <algorithm>
#include <map>
#define N 501
#define INF 2000000000
#define MOD 1000000007

using namespace std;
int n, r_len, L[N][N * 4];
long long dp[N][N * 4];
pair<int, int> in[N], r[N * 4];
set<int> S;
map<int, int> M;

void r_push(int x, int y) {
	if (x > y) return;
	r[++r_len] = { x,y };
	if (x == y) M[x] = r_len;
}
int main() {
	int path = 0;
	long long sum;

//	freopen("input.txt", "r", stdin);
	scanf("%d", &n);
	for (int i = 1; i <= n; i++) {
		scanf("%d %d", &in[i].first, &in[i].second);
		S.insert(in[i].first), S.insert(in[i].second);
	}

	for (auto i : S) r_push(path + 1, i - 1), r_push(i, i), path = i;
	for (int i = 1; i <= n; i++) in[i] = { M[in[i].first],M[in[i].second] };
		
	dp[0][0] = 1;
	for (int i = 1; i <= n; i++) {
		sum = dp[i][0] = 1;
		for (int j = 1; j <= r_len; j++) {
			dp[i][j] = dp[i - 1][j];
			if (in[i].first <= j && j <= in[i].second) {
				dp[i][j] += sum * (long long)(r[j].second - r[j].first + 1);
				if (!L[i - 1][j]) L[i][j] = i - 1;
				else {
					L[i][j] = L[i - 1][j];

					long long len_x = i - (long long)L[i][j], len_y = (long long)(r[j].second - r[j].first + 1);
					long long P = (len_x + len_y)*(len_x + len_y - 1) / 2 - (len_x + len_y - 1)*(len_x + len_y - 2) / 2;
					dp[i][j] += dp[L[i][j]][j - 1] * P;

				} dp[i][j] %= MOD;
			} sum = (sum + dp[i - 1][j]) % MOD;
		}
	}
	sum = 0;
	for (int i = 1; i <= r_len; i++) sum = (sum + dp[n][i]) % MOD;
	printf("%lld", sum);
}

Compilation message (stderr)

boat.cpp: In function 'int main()':
boat.cpp:26:17: warning: ignoring return value of 'int scanf(const char*, ...)', declared with attribute warn_unused_result [-Wunused-result]
  scanf("%d", &n);
                 ^
boat.cpp:28:46: warning: ignoring return value of 'int scanf(const char*, ...)', declared with attribute warn_unused_result [-Wunused-result]
   scanf("%d %d", &in[i].first, &in[i].second);
                                              ^

• Subtask 1: 9 / 9
• Subtask 2: 0 / 22
• Subtask 3: 0 / 27
• Subtask 4: 0 / 42