/ SeriousOJ /

Record Detail

Accepted


  
# Status Time Cost Memory Cost
#1 Accepted 1ms 532.0 KiB
#2 Accepted 15ms 588.0 KiB
#3 Accepted 47ms 556.0 KiB
#4 Accepted 20ms 532.0 KiB
#5 Accepted 22ms 532.0 KiB
#6 Accepted 20ms 548.0 KiB
#7 Accepted 20ms 532.0 KiB
#8 Accepted 42ms 532.0 KiB
#9 Accepted 44ms 532.0 KiB
#10 Accepted 20ms 532.0 KiB
#11 Accepted 20ms 532.0 KiB
#12 Accepted 20ms 532.0 KiB
#13 Accepted 38ms 532.0 KiB
#14 Accepted 20ms 532.0 KiB
#15 Accepted 20ms 532.0 KiB
#16 Accepted 20ms 532.0 KiB
#17 Accepted 21ms 532.0 KiB
#18 Accepted 45ms 532.0 KiB
#19 Accepted 20ms 532.0 KiB
#20 Accepted 20ms 532.0 KiB

Code

#include <iostream>
#include <cmath>
using namespace std;

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);

    int t;
    cin >> t;
    while(t--) {
        int n;
        cin >> n;
        long long sum = 0;
        for(int i = 0; i < n; i++) {
            int x;
            cin >> x;
            sum += x;
        }
        // Solve k*(k+1)/2 <= sum
        // Using quadratic formula: k = floor((-1 + sqrt(1 + 8*sum)) / 2)
        long long k = static_cast<long long>(floor((-1.0 + sqrt(1 + 8.0 * sum)) / 2));
        cout << k + 1 << "\n";
    }
    return 0;
}

Information

Submit By
Type
Submission
Problem
P1114 Maximize the MEX
Language
C++17 (G++ 13.2.0)
Submit At
2025-02-21 17:35:58
Judged At
2025-02-21 17:35:58
Judged By
Score
100
Total Time
47ms
Peak Memory
588.0 KiB