/ SeriousOJ /

Record Detail

Accepted


  
# Status Time Cost Memory Cost
#1 Accepted 1ms 324.0 KiB
#2 Accepted 3ms 580.0 KiB
#3 Accepted 3ms 580.0 KiB
#4 Accepted 3ms 540.0 KiB
#5 Accepted 3ms 532.0 KiB
#6 Accepted 3ms 532.0 KiB
#7 Accepted 3ms 580.0 KiB
#8 Accepted 2ms 584.0 KiB
#9 Accepted 2ms 364.0 KiB
#10 Accepted 3ms 576.0 KiB
#11 Accepted 1ms 532.0 KiB
#12 Accepted 2ms 580.0 KiB
#13 Accepted 3ms 592.0 KiB
#14 Accepted 3ms 592.0 KiB
#15 Accepted 3ms 580.0 KiB

Code

// #pragma GCC optimize("O3,unroll-loops,Ofast")
// #pragma GCC target("avx2")
#include<bits/stdc++.h>
// #include <ext/pb_ds/assoc_container.hpp>
// #include <ext/pb_ds/tree_policy.hpp>
// using namespace __gnu_pbds;
#define int long long
#define endl '\n'

using namespace std;
using pii = pair<int, int>;
using tup = tuple<int, int, int>;

// mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
// template <class T> using ordered_set = tree<T, null_type,
//                          less<T>, rb_tree_tag, tree_order_statistics_node_update>;

const int inf = (1e15)+5;
const int mod = 1000000007;
const double eps = 1e-6;
const int N = 200001;

void preprocess() {}

void solve(int tc) {
	int n, k;
	cin >> n >> k;
	vector<string> v(n);
	for(string &s : v) cin >> s;

	k++;
	string ans;
	for(int i=0; i+k-1<n; i++) {
		string now;
		for(int j=0; j<k; j++)
			now += v[i+j];
		
		if(now.size() > ans.size()) ans = now;
		else if(now.size() == ans.size() && now > ans) ans = now;
	}

	cout << ans << endl;
}
    
int32_t main() {
    cin.tie(NULL)->sync_with_stdio(false);
    cout.precision(10);

    preprocess();

    int T = 1;
    cin >> T;

    for (int i = 1; i <= T; i++) {
        solve(i);
    }

    return 0;
}

Information

Submit By
Type
Submission
Problem
P1083 Number concatenation
Contest
Bangladesh 2.0
Language
C++20 (G++ 13.2.0)
Submit At
2024-08-16 15:55:32
Judged At
2024-10-03 13:29:28
Judged By
Score
100
Total Time
3ms
Peak Memory
592.0 KiB