이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#include <iostream>
#include <fstream>
#include <vector>
#include <cstring>
#include <algorithm>
#include <set>
#include <map>
#include <queue>
#include <list>
#include <utility>
#include <cmath>
#include <numeric>
using namespace std;
typedef long long ll;
#define F first
#define S second
#define pb push_back
#define endl "\n"
#define all(x) x.begin(), x.end()
const int M = 200007;
const ll inf = 1e18;
const ll mod = 1e9 + 7;
const double pi = acos(-1);
const int dx[] = {1, 0, -1, 0}, dy[] = {0, 1, 0, -1};
int n, k, a[M];
int solve(int sum)
{
int l = 1, r = n + k, ret = 0;
while(l < r){
if(a[l] + a[r] == sum){
++l;
--r;
}
else if(a[l] + a[r] > sum){
--r;
if(ret != 0) ret = -1;
else ret = r + 1;
}
else{
++l;
if(ret != 0) ret = -1;
else ret = l - 1;
}
}
if(ret == 0) ret = (n + k - 1) / 2 + 1;
return ret;
}
int main()
{
ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
cin >> n >> k;
for(int i = 1; i <= n + k; ++i) cin >> a[i];
sort(a + 1, a + n + k + 1);
int res = solve(a[1] + a[n + k]);
if(res == -1) res = solve(a[2] + a[n + k]);
if(res == -1) res = solve(a[1] + a[n + k - 1]);
if(res == -1) res = solve(a[2] + a[n + k - 1]);
for(int i = 1; i <= n + k; ++i) if(i != res) cout << a[i] << " ";
cout << endl;
return 0;
}
// Don't forget special cases. (n = 1?)
// Look for the constraints. (Runtime array? overflow?)
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