제출 #31964

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
31964		minimario	수열 (APIO14_sequence)	C++11	0 / 100	0 ms	1840 KiB

sequence

#include <bits/stdc++.h>
using namespace std;
 
const int MAX = 1000005;
typedef pair<int, int> pii;
 
#define f first
#define s second
 
 
pii line[MAX];
int orig[MAX];
double hit[MAX];
int ct = 0;
 
double sect(pii l1, pii l2) {
	// l1.f * x + l1.s = l2.f * x + l2.s
	return (double)(l2.s - l1.s) / (double)(l1.f - l2.f);
 
}
void add(int m, int b, int k) {
	pii l = {m, b};
	// l and line[2] > hit[2]
	while (ct > 1 && sect(l, line[ct]) < hit[ct]) {
		ct--;
	}
	ct++;
	orig[ct] = k;
	line[ct] = l;
	if (ct != 1) { hit[ct] = sect(l, line[ct-1]); }
}
 
pii eval(int x) {
	pii l; int ind = -1;
	if (ct == 0) { return {0, 0}; }
	if (ct == 1) { 
		l = line[1]; 
		ind = orig[1];
	}
	else { 
		l = line[lower_bound(hit+2, hit+ct+1, (double)x) - hit - 1]; 
		ind = orig[lower_bound(hit+2, hit+ct+1, (double)x) - hit - 1];
	}
	return {l.f * x + l.s, ind};
 
}
 
int a[MAX];
int p[MAX];
int dp[MAX][205];
int last[MAX][205];
int main() {
	//freopen("a.in", "r", stdin);
	//freopen("a.out", "w", stdout);
	int n, k; scanf("%d %d", &n, &k);
	for (int i=1; i<=n; i++) {
		scanf("%d", &a[i]); 
		p[i] = p[i-1] + a[i];
	}
	for (int k=1; k<=n; k++) {
		ct = 0;
		for (int i=k-1; i<=n; i++) {
			if (k == 1) { 
				dp[i][k] = 0;
			}
			else {
				add(-p[i], -dp[i][k-1]+p[i]*p[i], i);
				if (i != k-1) { 
					dp[i][k] = -eval(p[i]).f; 
					last[i][k] = eval(p[i]).s;
				}
			}
		}
	}
	k++;
	printf("%d\n", dp[n][k]);
	while (k != 1) {
		int bef = last[n][k];
		printf("%d ", bef);
		n = bef; k--;
	}
 
}

컴파일 시 표준 에러 (stderr) 메시지

sequence.cpp: In function 'int main()':
sequence.cpp:55:34: warning: ignoring return value of 'int scanf(const char*, ...)', declared with attribute warn_unused_result [-Wunused-result]
  int n, k; scanf("%d %d", &n, &k);
                                  ^
sequence.cpp:57:21: warning: ignoring return value of 'int scanf(const char*, ...)', declared with attribute warn_unused_result [-Wunused-result]
   scanf("%d", &a[i]); 
                     ^

• Subtask 1: 0 / 11
• Subtask 2: 0 / 11
• Subtask 3: 0 / 11
• Subtask 4: 0 / 17
• Subtask 5: 0 / 21
• Subtask 6: 0 / 29