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Progressive hints first, then the full explanation and implementation when you're ready to cash out.
Review status
AI-generated and still unreviewed. Double-check the details before internalizing them.
Hints
Open only as much as you need to keep the solve alive.
There is no input. This is an April Fools special, so the whole “algorithm” is in the picture. Treat each cell as an object .
Ignore rows and columns at first. Compare arrows as actual drawings: direction, fill, and shape. You want arrows to become identical pairs plus leftover.
The graph matchings tag is basically trolling, but it helps: connect two cells if their arrows are identical. The odd one is the unmatched vertex.
The duplicate pairs are -, -, -, -, -, -, and -.
Those pairs cover cells. The only cell left is , so the accepted program literally prints C2. Absolute peak computing.
There is no input. The statement gives a picture and asks for exactly two characters, so this is a fixed-output problem. The title is literal: find the arrow that is the odd one out.
There are cells in the grid:
So a very suspicious structure is: seven identical pairs and one lonely arrow. Compare arrows by their exact drawing, not by some row/column formula. The pairs are:
These pairs cover arrows. The only unmatched cell is:
That is the required output.
The checker expects the coordinate of the unique arrow that appears once. Since the problem has no input, every valid solution is just a constant-output program.
This matches the official Codeforces tutorial for 2214A: https://codeforces.com/blog/entry/152573
The program does no computation depending on input.
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
void setIO() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
}
int main() {
setIO();
cout << "C2";
}#include <bits/stdc++.h>
using namespace std;
using ll = long long;
void setIO() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
}
int main() {
setIO();
cout << "C2";
}