結果
| 問題 |
No.3114 0→1
|
| コンテスト | |
| ユーザー |
|
| 提出日時 | 2025-05-21 15:54:30 |
| 言語 | C++23 (gcc 13.3.0 + boost 1.87.0) |
| 結果 |
AC
|
| 実行時間 | 4 ms / 2,000 ms |
| コード長 | 1,850 bytes |
| コンパイル時間 | 1,109 ms |
| コンパイル使用メモリ | 99,908 KB |
| 実行使用メモリ | 7,844 KB |
| 最終ジャッジ日時 | 2025-05-21 15:54:32 |
| 合計ジャッジ時間 | 2,475 ms |
|
ジャッジサーバーID (参考情報) |
judge3 / judge5 |
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| ファイルパターン | 結果 |
|---|---|
| sample | AC * 3 |
| other | AC * 30 |
ソースコード
//by gemini 2.5Pro (流石にFlashに比べて賢そう?)
#include <iostream>
#include <vector>
#include <string>
#include <algorithm> // Not strictly needed for this solution but common include
int main() {
// Optimize C++ standard streams for faster I/O.
std::ios_base::sync_with_stdio(false);
std::cin.tie(NULL);
int N;
std::cin >> N;
std::string s_str;
std::cin >> s_str;
// Convert the input string to a vector of characters for easy modification.
std::vector<char> s(s_str.begin(), s_str.end());
int operations = 0;
// Iterate through the string from left to right.
for (int i = 0; i < N; ++i) {
// We only care if the current character s[i] is '0', as it might complete a bad pattern.
if (s[i] == '0') {
// Check for the "010" pattern ending at index i.
// This pattern is s[i-2]s[i-1]s[i] == '0''1''0'.
// For this, s[i] is '0', s[i-1] must be '1', and s[i-2] must be '0'.
if (i >= 2 && s[i-1] == '1' && s[i-2] == '0') {
s[i] = '1'; // Change s[i] to '1' to fix the pattern.
operations++; // Increment the operation count.
}
// Else, if s[i] was not part of a "010" pattern (and thus not changed),
// check for the "00" pattern ending at index i.
// This pattern is s[i-1]s[i] == '0''0'.
// For this, s[i] is '0', and s[i-1] must be '0'.
// The 'else if' ensures this check only happens if the "010" fix was not applied to s[i].
else if (i >= 1 && s[i-1] == '0') {
s[i] = '1'; // Change s[i] to '1' to fix the pattern.
operations++; // Increment the operation count.
}
}
}
std::cout << operations << std::endl;
return 0;
}