結果

問題 No.1802 Range Score Query for Bracket Sequence
ユーザー tenten
提出日時 2022-01-12 17:26:43
言語 Java
(openjdk 23)
結果
AC  
実行時間 628 ms / 2,000 ms
コード長 3,024 bytes
コンパイル時間 2,680 ms
コンパイル使用メモリ 80,168 KB
実行使用メモリ 65,396 KB
最終ジャッジ日時 2024-06-30 02:21:45
合計ジャッジ時間 13,414 ms
ジャッジサーバーID
(参考情報)
judge3 / judge4
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 * 1
other AC * 24
権限があれば一括ダウンロードができます

ソースコード

diff #
プレゼンテーションモードにする

import java.io.*;
import java.util.*;
public class Main {
public static void main(String[] args) throws Exception {
Scanner sc = new Scanner();
int n = sc.nextInt();
int q = sc.nextInt();
char[] parens = ("X" + sc.next() + "X").toCharArray();
BinaryIndexedTree bit = new BinaryIndexedTree(n + 2);
for (int i = 1; i <= n; i++) {
if (parens[i] == ')' && parens[i - 1] == '(') {
bit.add(i, 1);
}
}
StringBuilder sb = new StringBuilder();
for (int i = 0; i < q; i++) {
int type = sc.nextInt();
if (type == 1) {
int idx = sc.nextInt();
if (parens[idx] == '(') {
parens[idx] = ')';
if (parens[idx - 1] == '(') {
bit.add(idx, 1);
}
if (parens[idx + 1] == ')') {
bit.add(idx + 1, -1);
}
} else {
parens[idx] = '(';
if (parens[idx + 1] == ')') {
bit.add(idx + 1, 1);
}
if (parens[idx - 1] == '(') {
bit.add(idx, -1);
}
}
} else {
int left = sc.nextInt();
int right = sc.nextInt();
sb.append(bit.getSum(left + 1, right)).append("\n");
}
}
System.out.print(sb);
}
}
class BinaryIndexedTree {
int size;
int[] tree;
public BinaryIndexedTree(int size) {
this.size = size;
tree = new int[size];
}
public void add(int idx, int value) {
int mask = 1;
while (idx < size) {
if ((idx & mask) != 0) {
tree[idx] += value;
idx += mask;
}
mask <<= 1;
}
}
public int getSum(int from, int to) {
return getSum(to) - getSum(from - 1);
}
public int getSum(int x) {
int mask = 1;
int ans = 0;
while (x > 0) {
if ((x & mask) != 0) {
ans += tree[x];
x -= mask;
}
mask <<= 1;
}
return ans;
}
}
class Scanner {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
StringTokenizer st = new StringTokenizer("");
public Scanner() throws Exception {
}
public int nextInt() throws Exception {
return Integer.parseInt(next());
}
public long nextLong() throws Exception {
return Long.parseLong(next());
}
public double nextDouble() throws Exception {
return Double.parseDouble(next());
}
public String next() throws Exception {
if (!st.hasMoreTokens()) {
st = new StringTokenizer(br.readLine());
}
return st.nextToken();
}
}
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