PROBLEM LINK:
Contest - Division 4
Contest - Division 3
Contest - Division 2
Contest - Division 1
DIFFICULTY:
EASY
PROBLEM:
Given an array A of N integers. You may perform atmost K moves of the following type - change the value of A_i to either of it’s adjacent values.
Determine the maximum possible value of \min (A_1,A_2,\dots,A_N).
EXPLANATION:
The solution of the problem boils down to one crucial observation - setting the value of A_i to either of it’s adjacent values is equivalent to erasing element A_i from A.
Then, the problem is reduced to erasing at most K elements from A such that the minimum value is maximised. This can be done by sorting A and erasing the first K smallest elements.
TIME COMPLEXITY:
O(N\log N)
per test case.
SOLUTIONS:
Editorialist’s solution can be found here.
Author's solution
#ifdef WTSH
#include <wtsh.h>
#else
#include <bits/stdc++.h>
using namespace std;
#define dbg(...)
#endif
#define int long long
#define endl "\n"
#define sz(w) (int)(w.size())
using pii = pair<int, int>;
const long long INF = 1e18;
const int N = 1e6 + 5;
// -------------------- Input Checker Start --------------------
long long readInt(long long l, long long r, char endd)
{
long long x = 0;
int cnt = 0, fi = -1;
bool is_neg = false;
while(true)
{
char g = getchar();
if(g == '-')
{
assert(fi == -1);
is_neg = true;
continue;
}
if('0' <= g && g <= '9')
{
x *= 10;
x += g - '0';
if(cnt == 0)
fi = g - '0';
cnt++;
assert(fi != 0 || cnt == 1);
assert(fi != 0 || is_neg == false);
assert(!(cnt > 19 || (cnt == 19 && fi > 1)));
}
else if(g == endd)
{
if(is_neg)
x = -x;
if(!(l <= x && x <= r))
{
cerr << l << ' ' << r << ' ' << x << '\n';
assert(false);
}
return x;
}
else
{
assert(false);
}
}
}
string readString(int l, int r, char endd)
{
string ret = "";
int cnt = 0;
while(true)
{
char g = getchar();
assert(g != -1);
if(g == endd)
break;
cnt++;
ret += g;
}
assert(l <= cnt && cnt <= r);
return ret;
}
long long readIntSp(long long l, long long r) { return readInt(l, r, ' '); }
long long readIntLn(long long l, long long r) { return readInt(l, r, '\n'); }
string readStringLn(int l, int r) { return readString(l, r, '\n'); }
string readStringSp(int l, int r) { return readString(l, r, ' '); }
void readEOF() { assert(getchar() == EOF); }
vector<int> readVectorInt(int n, long long l, long long r)
{
vector<int> a(n);
for(int i = 0; i < n - 1; i++)
a[i] = readIntSp(l, r);
a[n - 1] = readIntLn(l, r);
return a;
}
// -------------------- Input Checker End --------------------
int sumN = 0;
void solve()
{
int n = readIntSp(2, 1e5);
int k = readIntLn(0, 1e5);
sumN += n;
vector<int> a = readVectorInt(n, 1, 1e9);
sort(a.begin(), a.end());
k = min(k, n - 1);
cout << a[k] << endl;
}
int32_t main()
{
ios::sync_with_stdio(0);
cin.tie(0);
int T = readIntLn(1, 100);
for(int tc = 1; tc <= T; tc++)
{
// cout << "Case #" << tc << ": ";
solve();
}
readEOF();
assert(sumN <= 2e5);
return 0;
}
Tester's solution
// Super Idol的笑容
// 都没你的甜
// 八月正午的阳光
// 都没你耀眼
// 热爱105°C的你
// 滴滴清纯的蒸馏水
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/rope>
using namespace std;
using namespace __gnu_pbds;
using namespace __gnu_cxx;
#define int long long
#define ll long long
#define ii pair<ll,ll>
#define iii pair<ii,ll>
#define fi first
#define se second
#define endl '\n'
#define debug(x) cout << #x << ": " << x << endl
#define pub push_back
#define pob pop_back
#define puf push_front
#define pof pop_front
#define lb lower_bound
#define ub upper_bound
#define rep(x,start,end) for(auto x=(start)-((start)>(end));x!=(end)-((start)>(end));((start)<(end)?x++:x--))
#define all(x) (x).begin(),(x).end()
#define sz(x) (int)(x).size()
#define indexed_set tree<ll,null_type,less<ll>,rb_tree_tag,tree_order_statistics_node_update>
//change less to less_equal for non distinct pbds, but erase will bug
mt19937 rng(chrono::system_clock::now().time_since_epoch().count());
int n,k;
int arr[200005];
signed main(){
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
cin.exceptions(ios::badbit | ios::failbit);
int TC;
cin>>TC;
while (TC--){
cin>>n>>k;
rep(x,0,n) cin>>arr[x];
sort(arr,arr+n);
cout<<arr[min(k,n-1)]<<endl;
}
}