結果

問題 No.1717 Levi-Civita Triangle
ユーザー nok0nok0
提出日時 2021-08-04 23:12:44
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
TLE  
実行時間 -
コード長 1,185 bytes
コンパイル時間 2,987 ms
コンパイル使用メモリ 228,324 KB
最終ジャッジ日時 2025-01-23 14:08:57
ジャッジサーバーID
(参考情報)
judge4 / judge3
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 3
other AC * 26 TLE * 16
権限があれば一括ダウンロードができます
コンパイルメッセージ
main.cpp: In function ‘int main()’:
main.cpp:11:14: warning: ignoring return value of ‘int scanf(const char*, ...)’ declared with attribute ‘warn_unused_result’ [-Wunused-result]
   11 |         scanf("%d", &N);
      |         ~~~~~^~~~~~~~~~
main.cpp:17:22: warning: ignoring return value of ‘int scanf(const char*, ...)’ declared with attribute ‘warn_unused_result’ [-Wunused-result]
   17 |                 scanf("%d", &a);
      |                 ~~~~~^~~~~~~~~~

ソースコード

diff #
プレゼンテーションモードにする

#pragma GCC target("avx2")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
#include <bits/stdc++.h>
using namespace std;
int main() {
int N;
scanf("%d", &N);
bitset<200010> bit_0, bit_1, bit_2;
int a;
for(int i = 0; i < 2 * N + 1; i++) {
scanf("%d", &a);
// cout << a << endl;
if(a == 0) {
bit_0.set(i);
} else if(a == 1) {
bit_1.set(i);
} else {
bit_2.set(i);
}
}
// cout << -1 << " 0 " << bit_0.any() << endl;
// cout << -1 << " 1 " << bit_1.any() << endl;
// cout << -1 << " 2 " << bit_2.any() << endl;
bitset<200010> new_bit_0, new_bit_1, new_bit_2, bit_01, bit_10;
for(int i = 0; i < N; i++) {
bit_01 = (bit_0 & (bit_1 >> 1));
new_bit_1 = (bit_01 & (bit_2 >> 2)) |
(bit_1 & (bit_2 >> 1) & (bit_0 >> 2)) |
(bit_2 & (bit_01 >> 1));
bit_10 = (bit_1 & (bit_0 >> 1));
new_bit_2 = (bit_0 & (bit_2 >> 1) & (bit_1 >> 2)) |
(bit_2 & (bit_10 >> 1)) |
(bit_10 & (bit_2 >> 2));
swap(bit_1, new_bit_1);
swap(bit_2, new_bit_2);
bit_0 = (bit_1 ^ bit_2).flip();
}
if(bit_1.any())
putchar('1');
else if(bit_2.any())
putchar('2');
else
putchar('0');
return 0;
}
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