제출 #32106

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
32106		minimario	수열 (APIO14_sequence)	C++14	100 / 100	519 ms	86784 KiB

sequence

#include <bits/stdc++.h>
using namespace std;
 
const int MAX = 100005;
typedef long long ll;
typedef pair<ll, ll> pll;
 
 
#define f first
#define s second
 
 
pll line[MAX];
int orig[MAX];
int ct = 0;
int ptr = 0;
bool bad(pll l0, pll l1, pll l2) {
	return (l1.s-l0.s)*(l0.f-l2.f) >= (l0.f-l1.f)*(l2.s-l0.s);
}
 
ll f(pll l, ll x) {
	return l.f * x + l.s;
}
void add(pll l, int i) {
	while (ct >= 2 && ptr < ct && bad(line[ct-1], line[ct], l)) {
		ct--;
	}
	ptr = min(ptr, ct);
	ct++;
	line[ct] = l;
	orig[ct] = i;
}
 
pair<ll, int> get(ll x) {
	ptr = max(ptr, 1);
	if (ct == 0) { return {0, 0}; }
	while (ptr < ct && f(line[ptr], x) < f(line[ptr+1], x)) {
		ptr++;
	}
	return {f(line[ptr], x), orig[ptr]};
}
 
int a[MAX];
ll p[MAX];
ll dp[2][MAX];
int last[205][MAX];
int main() {
	// double c = clock();
	//freopen("a.in", "r", stdin);
	//freopen("a.out", "w", stdout);
 
	int n, k0; scanf("%d %d", &n, &k0);
	for (int i=1; i<=n; i++) {
		scanf("%d", &a[i]); 
		p[i] = (ll)p[i-1] + (ll)a[i];
	}
	for (int k=1; k<=k0+1; k++) {
		ct = 0;
		for (int i=k-1; i<=n; i++) {
			if (k == 1) { 
				dp[k][i] = 0;
			}
			else {
				add({p[i], (ll)dp[(k+1)%2][i]-(ll)p[i]*(ll)p[i]}, i);
				if (i != k-1) { 
					pll ans = get(p[i]);
					dp[k%2][i] = ans.f; 
					last[k][i] = (int)ans.s;
				}
			}
		}
	}
	k0++;
	printf("%lld\n", dp[k0%2][n]);
	while (k0 != 1) {
		int bef = last[k0][n];
		printf("%d ", bef);
		n = bef; k0--;
	}
	// cout << (clock() - c) / (double)CLOCKS_PER_SEC;
}

컴파일 시 표준 에러 (stderr) 메시지

sequence.cpp: In function 'int main()':
sequence.cpp:52:36: warning: ignoring return value of 'int scanf(const char*, ...)', declared with attribute warn_unused_result [-Wunused-result]
  int n, k0; scanf("%d %d", &n, &k0);
                                    ^
sequence.cpp:54:21: warning: ignoring return value of 'int scanf(const char*, ...)', declared with attribute warn_unused_result [-Wunused-result]
   scanf("%d", &a[i]); 
                     ^

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