Codeforces · 2195A

Sieve of Erato67henes

This C++ solution uses mathematical reasoning for Codeforces 2195A Sieve of Erato67henes. Read the reasoning, inspect the code, or try your own test case below.

2195AMathMathematical reasoningC++22 lines
Solution189of 248
Open official problem ↗ Download C++ file ↓ Search the library → Open full Code Lab ↗ Report an issue ↗

Approach

Mathematical reasoning

Sieve of Erato-67-henes and I/O match; because 67 is prime, the code's presence test is equivalent to the required condition.

Math

Problem and code

Useful links.

Written by benbenyaojifen. Try the problem first, then compare your approach with the code.

Open official problem ↗View exact source file ↗
Implementation

A_Sieve_of_Erato_67_henes.cpp

C++

    #include <bits/stdc++.h>
    using namespace std;
    using ll = long long;
    const int inf = 1e9;
    const long long INF = 1e17; //❄️
    void solve() {
        int n; cin >> n;
        vector<int> v(n);
        for (int i = 0; i < n; i++) {
            cin >> v[i];
        }
        if (find(v.begin(), v.end(), 67) != v.end()) cout << "Yes" << '\n';
        else cout << "No" << '\n';
    }
    int main() {
        ios::sync_with_stdio(0); cin.tie(0); 
        int t; cin >> t; 
        while (t--) {
            solve();
        }
        return 0;
    }
        

Did this explanation save you time? I'm a Grade 11 student building this free library to make difficult algorithms easier to understand.

Buy me a coffee ↗Keep studying →

Test this problem

Run your code here.

Paste your code, run a test case, compare the output, or trace selected values.

Full trace, comparison & stress testing ↗
StatusReady
Output
No run yet.
Diagnostics
No diagnostics yet.

Each run is isolated and has strict limits. Passing one test does not guarantee the judge will accept the solution.