結果

問題 No.1589 Bit Vector
コンテスト
ユーザー e869120
提出日時 2021-06-26 22:08:00
言語 C++14
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 14 ms / 2,000 ms
コード長 1,789 bytes
コンパイル時間 827 ms
コンパイル使用メモリ 80,628 KB
実行使用メモリ 6,944 KB
最終ジャッジ日時 2024-07-01 12:35:28
合計ジャッジ時間 3,068 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 * 1
other AC * 35
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ソースコード

diff # 

#include <iostream>
#include <vector>
#include <string>
#include <tuple>
using namespace std;

int N, K, T;
int B[1009], C[1009][1009];
vector<tuple<string, int, int, int>> Answer;

void Solve(int px, int py) {
	Answer.push_back(make_tuple("UPD", N, 1, -1));
	Answer.push_back(make_tuple("XOR", N, py, N));
	Answer.push_back(make_tuple("AND", N, px, N));
	Answer.push_back(make_tuple("XOR", px, px, N));
	Answer.push_back(make_tuple("XOR", py, py, N));
}

int main() {
	// Step #1. Input
	cin >> N >> K >> T;
	for (int i = 1; i <= T; i++) {
		for (int j = 0; j < N; j++) cin >> C[i][j];
	}

	// Step #2. Solve
	for (int i = N - 2; i >= 0; i--) {
		for (int j = 0; j <= i; j++) {
			Solve(j, j + 1);
		}
	}
	Answer.push_back(make_tuple("UPD", N, 0, -1));
	Answer.push_back(make_tuple("XOR", N, N - K, N));

	// Step #3. Output The Answer
	cout << Answer.size() << endl;
	for (int i = 0; i < Answer.size(); i++) {
		if (get<0>(Answer[i]) == "UPD") cout << get<0>(Answer[i]) << " " << get<1>(Answer[i]) << " " << get<2>(Answer[i]) << endl;
		if (get<0>(Answer[i]) == "AND") cout << get<0>(Answer[i]) << " " << get<1>(Answer[i]) << " " << get<2>(Answer[i]) << " " << get<3>(Answer[i]) << endl;
		if (get<0>(Answer[i]) == "XOR") cout << get<0>(Answer[i]) << " " << get<1>(Answer[i]) << " " << get<2>(Answer[i]) << " " << get<3>(Answer[i]) << endl;
	}

	/*for (int i = 1; i <= T; i++) {
		for (int j = 0; j < N; j++) B[j] = C[i][j]; B[N] = 0;
		for (int j = 0; j < Answer.size(); j++) {
			if (get<0>(Answer[j]) == "UPD") B[get<1>(Answer[j])] = get<2>(Answer[j]);
			if (get<0>(Answer[j]) == "AND") B[get<1>(Answer[j])] = (B[get<2>(Answer[j])] & B[get<3>(Answer[j])]);
			if (get<0>(Answer[j]) == "XOR") B[get<1>(Answer[j])] = (B[get<2>(Answer[j])] ^ B[get<3>(Answer[j])]);
		}
	}*/
	return 0;
}
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