Submission #998103

#	Time	Username	Problem	Language	Result	Execution time	Memory
998103		lovrot	Sparklers (JOI17_sparklers)	C++17	100 / 100	27 ms	2560 KiB

sparklers

#include <cstdio> 
#include <algorithm>
#include <cstring>

using namespace std; 

typedef long long ll;

const int N = 1e5 + 10;
const int OO = 1e9 + 10;

int n, t, k, A[N];

ll B[N];

bool out(int l, int r, int lo, int hi) { // lo <= l | r <= hi
	int l_ = l, r_ = r;

	for(; lo >= 0 && B[lo] >= B[r]; --lo) {
		if(B[lo] > B[l_]) {
			l_ = lo; 
		}
	}

	for(; hi < n && B[hi] <= B[l]; ++hi) {
		if(B[hi] < B[r_]) { 
			r_ = hi;
		}
	}

	if(lo < 0 && hi >= n) { return 1; }
	if(l == l_ && r == r_) { return 0; }
	return out(l_, r_, lo, hi); 
}

bool in(int l, int r, int lo, int hi) { // lo <= l | r <= hi
	int l_ = l, r_ = r;

	for(; lo < k && B[lo] >= B[r]; ++lo) {
		if(B[lo] > B[l_]) {
			l_ = lo; 
		}
	}

	for(; hi > k - 2 && B[hi] <= B[l]; --hi) {
		if(B[hi] < B[r_]) { 
			r_ = hi;
		}
	}

	if(lo >= k && hi <= k - 2) { return 1; }
	if(l == l_ && r == r_) { return 0; }
	return in(l_, r_, lo, hi); 
}

bool loop(int mi) {
	for(int i = 0; i < n; ++i) {
		B[i] = (ll) A[i] - (ll) 2 * mi * t * i;
	}	

	return out(k - 1, k - 1, k - 2, k) & in(0, n - 1, 1, n - 2);
}

int main() { 
	scanf("%d%d%d", &n, &k, &t);
	for(int i = 0; i < n; ++i) { scanf("%d", A + i); }

	int lo = -1, hi = OO / t + 100;
	for(int mi = (lo + hi) / 2; hi - lo > 1; mi = (lo + hi) / 2) {
		bool res = loop(mi);
//		printf("[%d %d %d] = %d\n", lo, mi, hi, res);
		if(res) { hi = mi; }
		else { lo = mi; }
	}

	printf("%d\n", hi);
	return 0;
}

Compilation message (stderr)

sparklers.cpp: In function 'int main()':
sparklers.cpp:65:7: warning: ignoring return value of 'int scanf(const char*, ...)' declared with attribute 'warn_unused_result' [-Wunused-result]
   65 |  scanf("%d%d%d", &n, &k, &t);
      |  ~~~~~^~~~~~~~~~~~~~~~~~~~~~
sparklers.cpp:66:36: warning: ignoring return value of 'int scanf(const char*, ...)' declared with attribute 'warn_unused_result' [-Wunused-result]
   66 |  for(int i = 0; i < n; ++i) { scanf("%d", A + i); }
      |                               ~~~~~^~~~~~~~~~~~~

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