이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#include <iostream>
#include <vector>
#include <cmath>
#include <algorithm>
#include "books.h"
using namespace std;
typedef long long ll;
void solve(int N, int K, long long A, int S){
vector <ll> srch(N+1. -1);
ll suma = 0;
for (int i = 1; i <= min(N,K); i++){
srch[i] = skim(i);
suma += srch[i];
}
if (suma > 2*A)
impossible();
vector <int> odg;
if (suma >= A && suma <= 2*A){
for (int i = 1; i <= K; i++)
odg.push_back(i);
answer(odg);
}
int l = 1, r = N + 1;
while (l < r){
ll mid = (l+r)/2;
srch[mid] = skim(mid);
if (srch[mid] >= A)
r = mid;
else
l = mid + 1;
}
for (int i = min(N,l); i >= max(0, l - K); i--){
if (srch[i] == -1)
srch[i] = skim(i);
}
if (l > K && l <= N){
ll nsuma = suma - srch[K] + srch[l];
if (nsuma >= A && nsuma <= 2*A){
for (int i = 1; i <= K -1; i++)
odg.push_back(i);
odg.push_back(l);
answer(odg);
}
}
vector <bool> uklj(N + 1, false);
for (int i = 1; i <= K; i++)
uklj[i] = true;
for (int i = 1; i <= min(K, l - K - 1); i++){
suma -= srch[i];
suma += srch[max(K+i, l-K+i-1)], uklj[max(K+i, l-K+i-1)] = true, uklj[i] = false;
if (suma >= A && suma <= 2*A){
for (int i = 1; i <= N; i++)
if (uklj[i])
odg.push_back(i);
answer(odg);
}
}
impossible();
}
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